A  MATHEMATICAL 


SOLUTION  BOOK 


CONTAINING 


SYSTEMATIC  SOLUTIONS  OF  MANY  OF  THE 
MOST  DIFFICULT  PROBLEMS. 


Taken  from  the  Leading  Authors  on  Arithmetic  and  Algebra,  Many  Prob- 
lems and  Solutions  from   Geometry,  Trigonometry  and  Calculus, 
Many  Problems  and  Solutions  from  the  Leading  Math- 
ematical  Journals  of  the  United   States,  and 
Many  Original  Problems  and  Solutions. 


WITH 


NOTES  AND  EXPLANATIONS 


BY 

B.  F.  FINKEL,  A.  M.,  M.  Sc. 
ii 

Member  of  the  London   Mathematical   Society,  Member  of  the  American 

Mathematical  Society,  Editor  of  the  American  Mathematical 

Monthly,  and  Professor  of  Mathematics  and 

Physics  in  Drury  College. 


THIRD  EDITION- REVISED. 


KIBLER  &  COMPANY,  PUBLISHERS, 

Springfield,  Mo. 


COPYRIGHT,  1888, 

BY 

B.    F.    FINKEL, 

IN  THE   OFFICE   OF  THE   LIBRARIAN   OF 
CONGRESS, 

WASHINGTON,  D.  C. 


PREFACE. 


This  work  is  the  outgrowth  of  eight  years'  experience  in 
teaching  in  the  Public  Schools,  during  which  time  I  have  ob- 
served that  a  work  presenting  a  systematic  treatment  of  solutions 
of  problems  would  be  serviceable  to  both  teachers  and  pupils. 

It  is  not  intended  to  serve  as  a  key  to  any  work  on  mathe- 
matics ;  but  the  object  of  its  appearance  is  to  present,  for  use  in 
the  schoolroom,  such  an  accurate  and  logical  method  of  solving 
problems  as  will  best  awaken  the  latent  energies  of  pupils,  and 
teach  them  to  be  original  investigators  in  the  various  branches  of 
science. 

It  will  not  be  denied  by  any  intelligent  educator  that  the  so- 
called  "Short  Cuts"  and  "Lightning  Methods"  are  positively  in- 
jurious to  beginners  in  mathematics.  All  the  "whys"  are  cut 
out  by  these  methods  and  the  student  robbed  of  the  very  object 
for  which  he  is  studying  mathematics ;  viz.,  the  devolpment  of 
the  reasoning  faculty  and  the  power  to  express  his  thoughts  in 
a  forcible  and  logical  manner.  By  pursuing  these  methods, 
mathematics  is  made  a  mere  memory  drill  and  when  the  memory 
fails,  all  is  lost ;  whereas,  it  should  be  presented  in  such  a  way  as 
to  develop  the  memory,  the  imagination,  and  the  reasoning  fac- 
ulty. By  following  out  the  method  pursued  in  this  book,  the 
mind  will  be  strengthened  in  these  three  powers,  besides  a  taste 
for  neatness  and  a  love  of  the  beautiful  will  be  cultivated. 

Any  one  who  can  write  out  systematic  solutions  of  problems 
can  resort  to  "Short  Cuts"  at  pleasure ;  but,  on  the  other  hand, 
let  a  student  who  has  done  all  his  work  in  mathematics  by  form- 
ulae, "Short  Cuts,"  and  "Lightning  Methods"  attempt  to  write 
out  a  systematic  solution  —  one  in  which  the  work  explains 
itself  —  and  he  will  soon  convince  one  of  his  inability  to  express 
his  thoughts  in  a  logical  manner.  These  so-called  "Short  Cuts" 
should  not  be  used  at  all,  in  the  schoolroom.  After  pupils  and 
students  have  been  drilled  on  the  systematic  method  of  solving 
problems,  they  will  be  able  to  solve  more  problems  by  short 
methods  than  they  could  by  having  been  instructed  in  all  the 
"Short  Cuts"  and  "Lightning  Methods"  extant. 

It  can  not  be  denied  that  more  time  is  given  to,  and  more 
time  wasted  in  the  study  of  arithmetic  in  the  public  schools  than 


2  PREFACE. 

in  any  other  branch  of  study ;  and  yet,  as  a  rule,  no  better  results 
are  obtained  in  this  branch  than  in  any  other.  The  reason  of 
this,  to  my  mind,  is  apparent.  Pupils  are  allowed  to  combine 
the  numbers  in  such  a  way  as  "to  get  the  answer"  and  that  is  all 
that  is  required.  They  are  not  required  to  tell  why  they  do  this, 
or  why  they  do  that,  but,  "did  you  get  the  answer?"  is  the 
question.  The  art  of  "ciphering"  is  thus  developed  at  the  ex- 
pense of  the  reasoning  faculty. 

The  method  of  solving  problems  pursued  in  this  book  is 
often  called  the  "Step  Method."  But  we  might,  with  equal  pro- 
priety, call  any  orderly  manner  of  doing  any  thing,  the  "Step 
Method."  There  are  only  two  methods  of  solving  problems — a 
right  method  and  a  wrong  method.  That  is  the  right  method 
which  takes  up,  in  logical  order,  link  by  link,  the  chain  of  rea- 
soning and  arrives  at  the  correct  result.  Any  other  method  is 
wrong  and  hurtful  when  pursued  by  those  who  are  beginners  in 
mathematics. 

One  solution,  thoroughly  analyzed  and  criticised  by  a  class, 
is  worth  more  than  a  dozen  solutions  the  difficulties  of  which  are 
seen  through  a  cloud  of  obscurities. 

This  book  can  be  used  to  a  great  advantage  in  the  class- 
room— the  problems  at  the  end  of  each  chapter  affording  ample 
exercise  for  supplementary  work. 

Many  of  the  Formulae  in  Mensuration  have  been  obtained  by 
the  aid  of  the  Calculus,  the  operation  alone  being  indicated.  This 
feature  of  the  work  will  not  detract  any  from  its  merits  for  those 
persons  who  do  not  understand  the  Calculus ;  for  those  who  do- 
understand  the  Calculus  it  will  afford  an  excellent  drill  to  work 
out  all  the  steps  taken  in  obtaining  the  formulae.  Many  of  the 
formulae  can  be  obtained  by  elementary  geometry  and  algebra. 
But  the  Calculus  has  been  used  for  the  sake  of  presenting  the 
beauty  and  accuracy  of  that  powerful  instrument  of  mathematics. 

In  cases  in  which  the  formulae  lead  to  series,  as  in  the  case 
of  the  circumference  of  the  ellipse,  the  rule  is  given  for  a  near 
approximation. 

It  has  been  the  aim  to  give  a  solution  of  every  problem 
presenting  anything  peculiar,  and  of  those  which  go  the  rounds 
of  the  country.  Any  which  have  been  omitted  will  receive  space 
in  future  editions  of  this  work.  The  limits  of  this  book  have 
compelled  me  to  omit  much  curious  and  valuable  matter  in 
Higher  Mathematics. 

I  have  taken  some  problems  and  solutions  from  the  School 
Visitor,  published  by  John  S.  Royer;  the  Mathematical  Maga- 
zine, and  the  Mathematical  Visitor,  published  by  Artemas  Mar- 
tin, A.  M.,  Ph.  D.,  LL.  D.;  and  the  Mathematical  Messenger, 
published  by  G.  H.  Harvill,  by  the  kind  permission  of  these 
distinguished  gentlemen. 


PREFACE.  3 

It  remains  to  acknowledge  my  indebtedness  to  Prof.  William 
Hoover,  A.  M.,  Ph.  D.,  of  the  Department  of  Mathematics  and 
Astronomy  in  the  Ohio  University  at  Athens,  for  critically  read- 
ing the  manuscript  of  the  part  treating  on  Mensuration. 

Hoping  that  the  work  will,  in  a  measure,  meet  the  object 
for  which  it  is  written,  I  respectfully  submit  it  to  the  use  of 
my  fellow  teachers  and  co-laborers  in  the  field  of  mathematics. 

Any  correction  or  suggestion  will  be  thankfully  received  by 
communicating  the  same  to  rne. 

THE  AUTHOR. 

1     - 

In  bringing  out  a  second  edition  of  this  work,  I  am  greatly 
indebted  to  Dr.  G.  B.  M.  Zerr  for  critically  reading  the  work 
with  a  view  to  eliminating  all  errors. 

THE  AUTHOR. 

Drury  College,  Feb.  Jp,  1897. 


PREFACE  TO  SECOND  EDITION. 


PREFACE  TO   THIRD    EDITION. 

The  hearty  reception  accorded  this  book,  as  is  attested  by 
the  fact  that  two  editions  of  1,200  copies  each  have  already  been 
sold,  encouraged  me  to  bring  out  this  third  edition. 

In  doing  so,  I  have  availed  myself  of  the  opportunity  of 
making  some  important  corrections,  and  such  changes  and  im- 
provements as  experience  and  the  suggestions  of  teachers  using 
the  book  have  dictated.  The  very  favorable  comments  on  the 
work  by  some  of  the  most  eminent  mathematicians  in  this 
country  confirm  the  opinion  that  the  book  is  a  safe  one  to 
put  into  the  hands  of  teachers  and  students. 

While  mathematics  is  the  exact  science,  yet  not  every  book 
that  is  written  upon  it  treats  of  it  as  though  it  were  such.  In- 
deed, until  quite  recently,  there  were  very  few  books  on  Arith- 
metic, Algebra,  Geometry  or  Calculus  that  were  not  mere  copies 
of  the  works  written  a  century  ago,  and  in  this  way  the  method, 
the  spirit,  the  errors  and  the  solecisms  of  the  past  two  hundred 
years  were  preserved  and  handed  down  to  the  present  genera- 
tion. At  the  present  time  the  writers  on  these  subjects  are 
breaking  away  from  the  beaten  paths  of  tradition,  and  the  re- 
sult, though  not  wholly  apparent,  is  a  healthier  and  more  vig- 
orous mathematical  philosophy.  Within  the  last  twenty-five 


4  PREFACE. 

years  there  has  set  in,  in  America,  a  reaction  against  the  spirit 
and  the  method  of  previous  generations,  so  that  C.  A.  Laisant, 
in  his  La  Mathematique  Phi  to  sop  hie  Enseignement :,  Paris,  1898, 
says,  "No  country  has  made  greater  progress  in  mathematics 
during  the  past  twenty-five  years  than  the  United  States.  The 
most  of  the  text-books  on  Arithmetic,  Algebra,  Geometry,  and 
the  Calculus,  written  within  the  last  five  years,  are  evidence  of 
this  progress. 

The  reaction  spoken  of  was  brought  about,  to  some  ex- 
tent, by  the  introduction  into  our-  higher  institutions  of  learn- 
ing of  courses  of  study  in  mathematics  bearing  on  the  wonder- 
ful researches  of  Abel,  Cauchy,  Galois,  Riemann,  Weirstrass, 
and  others.  This  reaction,  it  may  be  said,  started  as  early  as 
1832,  the  time  when  Benjamin  Peirce,  the  first  American  worthy 
to  be  ranked  with  Legendre,  Wallis,  Abel  and  the  Bernouillis, 
became  professor  of  mathematics  and  natural  philosophy  at 
Harvard  University.  Since  that  time  the  mathematical  courses 
in  our  leading  Universities  have  been  enlarged  and  strengthened 
until  now  the  opportunity  for  research  work  in  mathematics 
as  offered,  for  example,  at  the  University  of  Chicago,  Harvard, 
Yale,  Cornell,  Johns  Hopkins,  Princeton,  Columbia  and  others, 
is  as  good  as  is  to  be  found  anywhere  in  the  world.  For  ex- 
ample, the  following  are  the  subjects  offered  at  Harvard  for  the 
Academic  year  1899-1900:  Logarithms,  Plane  and  Spherical 
Trigonometry;  Plane  Analytical  Geometry;  Plane  and  Solid 
Analytical  Geometry ;  Algebra ;  Theory  of  Equations. — Invar- 
iants ;  Differential  and  Integral  Calculus ;  Modern  Methods  in 
Geometry. — Determinants ;  Elements  of  Mechanics ;  Quater- 
nions with  application  to  Geometry  and  Mechanics ;  Theory  of 
Curves  and  Surfaces;  Dynamics  of  a  Rigid  Body;  Trigonomet- 
ric Series. —  Introduction  to  Spherical  Harmonics. —  Potential 
Function  ;  Hydrostatics. — Hydrokinematics. — Force  Functions 
and  Velocity-Potential  Functions  and  their  uses. — Hydroki- 
netics ;  Infinite  Series  and  Products ;  The  Theory  of  Functions ; 
Albegra. — Galois's  Theory  of  Equations;  Lie's  Theory  as  ap- 
plied to  Differential  Equations ;  Riemann's  Theory  of  Func- 
tions ;  The  Calculus  of  Variations ;  Functions  Defined  by  Linear 
Differential  Equations ;  The  Theory  of  Numbers ;  The  Theory 
of  Planetary  Motions;  Theory  of  Surfaces;  Linear  Associative 
Algebra;  the  Algebra  of  Logic;  the  Plasticity  of  the  Earth; 
Elasticity;  and  the  Elliptic  and  the  Abelian  Transcendants. 

While  the  great  activity  and  real  progress  in  mathematics 
is  going  on  in  our  higher  institutions  of  learning,  a  like  degree 
of  activity  is  not  yet  being  manifested  in  many  of  our  colleges 
and  academies  and  the  Public  Schools  in  general.  It  is  not 
desirable  that  the  quantity  of  mathematics  studied  in  our  Public 
Schools  be  increased,  but  it  is  desirable  that  the  quality  of 


PREFACE.  5 

the  teaching  should  be  greatly  improved.  To  bring  about  this 
result  is  the  aim  of  this  book. 

It  does  not  follow,  as  is  too  often  supposed,  that  any  one 
familiar  with  the  multiplication  table,  and  able,  perhaps,  to 
solve  a  few  problems,  is  quite  competent  to  teach  Arithmetic, 
or  "Mathematics,"  as  arithmetic  is  popularly  called.  The  very 
first  principles  of  the  subject  are  of  the  utmost  importance, 
and  unless  the  correct  and  refined!  notions  of  these  principles 
are  presented  at  the  first,  quite  as  much  time  is  lost  by  the 
student  in  unlearning  and  freeing  himself  from  erroneous  con- 
ceptions as  was  required  in  acquiring  them.  Moreover,  no  ad- 
vance in  those  higher  modern  developments  in  Mathematics  is 
possible  by  any  one  having  false  notions  of  its  first  principles. 

As  a  branch  for  mental  discipline,  mathematics,  when 
properly  taught,  has  no  superior.  Other  subjects  there  are 
that  are  equally  beneficial,  but  none  superior.  The  idea  en- 
tertained by  many  teachers, — generally  those  who  have  pre- 
pared themselves  to  teach  other  subjects,  but  teach  mathematics 
until  an  opportunity  to  teach  in  their  special  line  presents  itself 
to  them, — that  mathematics  has  only  commercial  value  and 
only  so  much  of  it  should  be  studied  as  is  needed  by  the  student 
in  his  business  in  after  life,  is  pedagogically  and  psychologically 
wrong.  Mathematics  has  not  only  commercial  value,  but  edu- 
cational and  ethical  value  as  well,  and  that  to  a  degree  not 
excelled  by  any  other  science.  No  other  science  offers  such 
rich  opportunity  for  original  investigation  and  discovery.  So  far 
from  being  a  perfected  and  complete  body  of  doctrine  "handed 
down  from  heaven"  and  incapable  of  growth,  as  many  sup- 
pose, it  is  a  subject  which  is  being  developed  at  such  a  mar- 
velous rate  that  it  is  impossible  for  any  but  the  best  to  keep 
in  sight  of  its  ever-increasing  and  receding  boundary.  Because, 
therefore,  of  the  great  importance  of  mathematics  as  an  agent 
in  disciplining  and  developing  the  mind,  in  advancing  the  ma- 
terial comforts  of  man  by  its  application  in  every  department 
of  art  and  invention,  in  improving  ethical  ideas,  and  in  culti- 
vating a  love  for  the  good;  the  beautiful,  and  the  true,  the 
teachers  of  mathematics  should  have  the  best  training  possible. 
If  this  book  contributes  to  the  end,  that  a  more  comprehensive 
view  be  taken  of  mathematics,  better  services  rendered  in  pre- 
senting its  first  principles,  and  greater  interest  taken  in  its 
study,  I  shall  be  amply  rewarded  for  my  labor  in  its  prepa- 
ration. 

In  this  edition  I  have  added  a  chapter  on  Longitude  and 
Time,  the  biographies  of  a  few  more  mathematicians,  several 
hundred  more  problems  for  solution,  an  introduction  to  the 
study  of  Geometry,  and  an  introduction  to  the  study  of  Algebra. 

The  list  of  biographies  could  have  been  extended  indefi- 
nitely, but  the  student  who  becomes  interested  in  the  lives  of 


6  PREFACE. 

a  class  of  men  who  have  contributed  much  to  the  advancement 
of  civilization,  will  find  a  short  sketch  of  the  mathematicians 
from  the  earliest  times  down  to  the  present  day  in  Cajori's 
History  of  Mathematics  or  Ball's  A  Short  History  of  Mathematics. 

The  biographies  which  have  been  added  were  taken  from 
the  American  Mathematical  Mnothly.  I  have  received  much  aid 
in  my  remarks  on  Geometry  from  Study  and  Difficulties  of 
Mathematics,  by  Augustus  De  Morgan. 

It  yet  remains  for  me  to  express  my  thanks  to  my  colleague 
and  friend,  Prof.  F.  A.  Hall,  of  the  Department  of  Greek,  for 
making  corrections  in  the  Greek  terms  used  in  this  edition, 

THE  AUTHOR. 
Drury  College,  July,  1899. 


CONTENTS. 


CHAPTER  I. 

DEFINITIONS. 


Mathematics  classified 11  | 

CHAPTER  II. 
NUMERATION  AND  NOTATION. 


PAGE 

Definitions 11-14 


Numeration  defined 14 

French  Method  defined 14 

English  Method  defined 14 

Periods  of  Notation 15 


Notation  defined 

Arabic  Notation  defined 


15 
15 


Roman  Notation  defined 15 

Ordinal  Numbers 15 

Fractions 18 

Irrational  Numbers. 20 

Examples 21 


Addition  defined 


Subtraction  defined 


CHAPTER  III. 

ADDITION. 

22  |   Examples   23 

CHAPTER  IV. 
SUBTRACTION. 

23  |   Examples   24 

CHAPTER  V. 
MULTIPLICATION. 

Multiplication  defined 24  |  Examples   25-26 

CHAPTER  VI. 
DIVISION. 

Division  defined 26  |  Examples   27 

CHAPTER  VII. 
COMPOUND  NUMBERS. 


Definitions 28 

Time  Measure 29 

Definitions  in  Time  Measure .  23-31 

Longitude  and  Time 31-34 

Standard  Time 34-35 


The  International  Date  Line .  36-37 

Examples  37-36 

Solutions 38-39 

Examples   39-40 


Divisor  defined 41 

Common  Divisor  defined 41 


Multiple  defined 42 

Common  Multiple  defined 42 


CHAPTER  VIII. 
GREATEST  COMMON  DIVISOR. 

Greatest  Com'n  Divisor  defined.    41 
Examples   41-42 

CHAPTER  IX. 
LEAST  COMMON   MULTIPLE. 

Least  Common  Multiple  defined    42 


Examples   43-44 


8 


CONTENTS. 


CHAPTER  X. 
FRACTIONS. 

PAGB 


Definitions 44-46 

Fractions  classified 44 


Solutions  of  Problems. 
Examples       


PAGE 
46-49 
49-52 


CHAPTER  XI. 

CIRCULATING  DECIMALS. 


54 
55 
55 


IV. 


I.  Addition  of  Circulates 

II.  Subtraction  of  Circulates . . 
III.  Multiplication  of  Circulates 

CHAPTER  XII. 
PERCENTAGE. 


Division  of  Circulates  . 56 

Examples 56-57 


.Definitions .......       57 

Solutions 57-69 

II.    Commission  . 69 

Definitions  ......    ....     69 

Solutions   69-71 

Examples .     72 

Trade  Discount 72 

Definitions 72 

Solutions    73-76 

Examples 76 

Profit  and  Loss 77 


III. 


IV. 


Definitions 77 

Examples 83-84 

V.     Stocks  and  Bonds 84 

Definitions    .    . .       84 

Solutions 85-95 

Examples 96-97 

VI.    Insurance 97 

Definitions 97-98 

Solutions   . .    98-101 

Examples 102 


CHAPTER  XIII. 

INTEREST. 


II. 


III. 


Simple  Interest 103 

Definitions 103 

Solutions  103-106 

True  Discount  .  106 

Definitions  106 

Solutions  106-107 

Bank  Discount 107 

Definitions  . .  .107 


Solutions   108-109 

IV.    Annual  Interest 109 

Annual  Interest  defined  109 

Solutions    110-112 

V,     Compound  Interest 112 

Compound  Int.  defined.   112 
Solutions   113-115 


Definitions 
Solutions  . 


CHAPTER  XIV. 

ANNUITIES. 

. . . .   115  I  Examples 
.116-125 


126 


CHAPTER  XV. 

MISCELLANEOUS  PROBLEMS. 
Solutions .  127-139 


CHAPTER  XVI. 
RATIO  AND  PROPORTION. 

Definitions   139-141   I   Problems 

Solutions.    .........      141-144 


...  144-146 


Analysis  defined.  . . 
•Solutions  .  ...  .  . 


CHAPTER  XVII. 

ANALYSIS. 
. . . .   146      Problems 
146-177 


177-180 


CONTENTS. 


CHAPTER  XVIII. 

ALLIGATION. 


PAGE. 

I.  Alligation  Medial 181 

II.  Alligation  Alternate. . 181 


Solutions 


PAGE 

181-186 


CHAPTER  XIX. 
SYSTEMS  OF  NOTATION. 


Definitions 187 

Names  of  Systems 187 


Solutions 188-191 


CHAPTER  XX. 
MENSURATION. 


v 
V, 


Definitions 192-197 

Geometrical  Magnitudes  class- 
ified       192 

I.  Parallelogram  .  .  198-200 

II.  Triangles 200-204 

III.  Trapezoid    204-205 

IV.  Trapezium  and  Irregular 

Polygons 205 

V.  Regular  Polygons  . . .   205-207 

VI.  Circles   207-210 

VII.  Rectification      of     Plane 
Curves  and  Quadrature  of 

Plane  Surfaces 210-213 

II.  Conic  Sections     223 

Definitions 223-224 

1.  Ellipse 224-227 

2.  Parabola 227-229 

3.  Hyperbola    229-232 

IX.  Higher  Plane  Curves  ...     233 

1.  The  Cissoid  Diocles.   233-234 

2.  The  Conchoid  of  Nicom- 

edes  234-235 

3.  The  Oval  of  Cassini 235 

•    4.  The  Lemniscate  of   Ber- 

nouilli 236 

5.  The  Witch  of  Agnesi  236-237 

6.  The  Limacon 237 

7.  The  Quadratrix 238 

8.  The  Catenary 238-239 

9.  The  Tractrix 240 

10.  The  Syntractrix 240 

II.  Roulettes 240 

(a)  Cycloids 240-243 

(b)  Prolate    and    Curtate 
Cycloid 243-244 

(c)  Epitrochoid  and   Hy- 
potrochoid 245-248 

X.  Plane  Spiral 248 

1.  Spirals  of  Archimedes  249 

2.  The  Reciprocal  Spiral  249 

3.  The  Lituus 250 


4.  The  Logarithmic  Spi- 
ral   250 

XI.  Mensuration  of  Sol- 
ids     251-254 

1.  Cylinder 254-255 

2.  CylindricUngulas  255-262 

3.  Pyramid  and  Cone  262-266 

4.  Conical  Ungulas.   266-270 
XII.  Sphere 270-276 

XIII.  Spheroid    276-278 

1.  The  Prolate  Sphe- 

roid    276-278 

2.  The  Oblate  Sphe- 

roid    278-282 

XIV.  Conoids 282 

1.  The  Parbolic  Co- 

noid      ....  282-285 

2.  Hyperbolic  Co- 

noid    285-286 

XV.  Quadrature  and  Cuba- 
ture  of  Surfaces  and 
Solids  of  Revolution  286 

1.  Cycloid 286-287 

2.  Cissoid    287-288 

3.  Spindles  288-289 

4.  Parabolic  Spindle  289-290 
XVI.  Regular  Solids 290 

1.  Tetrahedron    ....   291-292 

2.  Octahedron    292* 

3.  Dodecahedron  .  . .   292-293 

4.  Icosahedron 293-294 

XVII.  Prismatoid 294-295 

XVIII.  Cylindric  Rings  ...  295-297 
XIX.  Miscellaneous    Measure- 
ments   297 

1.  Masons'     and 
Bricklayers'  work  297 

2.  Gauging 297-298 

3.  Lumber  Measure .  298 

4.  Grain  and  Hay.  .  ,   298-299 


10 


CONTENTS. 


MENSURATION  —  Concluded. 

PAGE 

XX.  Solutions  of  Miscellaneous  Problems 299-345 

Problems 346-354 

Examination  Tests : .  354-360 

Problems 361-366 

GEOMETRY. 

Definitions  367  (e)  Assumption  of  the  Sphere . .  380 

On  Geometric  Reasoning 369      (/)  of  Motion 380 

On  the  Advantages  Derived  from  On  Logic 380 

the  Study  of  Geometry  and  Laws  of  Thought 381 

Mathematics  in  General 370      Law  of  Converse 383 

Axioms 375      Methods  of  Reasoning 384 

General  Axioms  376  How  to  Prepare  a    Lesson  in 

Assumptions  .   377          Geometry 388 

(a)  Assumption  of  Straight  Line  377      Plane  Geometry 390 

(b)  of  the  Plane 377      The  Three  Famous  Problems  of 

(c)  of  Parallel  Line  377          Antiquity   408 

(d)  of  the  Circle...  379 

ALGEBRA. 

Definitions 415  Arithmetical  Fallacies 424 

Solutions  of  Problems 416-419  Probability 425 

The  Quadratic  Equation 419  Problems        428 

Indeterminate  forms 421 

Biography  of  Prof.  William  Hoover 403 

Probability  Problems 434-435 

Biography  of  Dr.  Artemas  Martin 436 

Biography  of  Prof.  E.  B.  Seitz 440 

Biography  of  Rene7  Descartes 442 

Biography  of  Leonhard  Euler 446 

Biography  of  Spphus  Lie 451 

Biography  of  Simon  Newcomb 454 

Biography  of  George  Bruce  Halsted  457 

Biography  of  Prof.  Felix  Klein 459 

Biography  of  Benjamin  Peirce 462 

Biography  of  James  Joseph  Sylvester s 468 

Biography  of  Arthur  Cayley 475 

Table  I 473 

II 478 

™ 478 

•JV 479 

V 479 

VI 480 

VII 480 

Example 4gl 


CHAPTER    I. 

DEFINITIONS. 

1.      Mathematics      (p.aftv)p.a.Ttxrj,      science)    is    that    science 
which  treats  of  quantity. 


!(1.)  Arithmetic. 
(2.)    Algebra... 

*.j  Geometry.. 


rl.  Calculus 

1 2.  Quaternions. 

El.  Platonic  Geometry.. 
2.  Analytical  Geometry. 
3.  Descriptive  Geometry. 


Differential. 

Integral. 

Calculus  of  Variations. 


: a.  Pure  Geometry.      » 
b.  Conic  Sections.  i\.  Plane  Trigon'y. 
c.  Trigonometry..  <2.  Analytical  Trig. 
(3.  Spherical      " 


'(1.)  Mensuration. 

(2.)  Surveying. 

(3.)  Navigation. 

(4.)  Mechanics. 

(5.)  Astronomy. 

(6.)  Optics. 

(7.)  Gunnery. 
^(8.)  &c.,  &c. 

2>.      Pure  Mathematics   treats  of  magnitude   or  quantity 
without  relation  to  matter. 

3.  Applied  Mathematics  treats  of  magnitude  as  subsist- 
ing in  material  bodies. 

4.  Arithmetic     (api^^nx^,      from     d^etf/io?,    a  number)    is 
the  science  of  numbers  and  the  art  of  computing  by  them. 

5.  Alyebra   (Ar.  al,    the,    and   geber,   philosopher)  is    that 
method  of  mathematical   computation  in  which   letters  and  other 
symbols  are  employed. 

6.  G-eOWietry      (y£ajfj.£Tpia,       from        y£U){j.£Tp£iv      to   measure 
land,  from     p£a,  yfh    the  earth,  and    p.£Tp£~iv,     to    measure)    is    the 
science  of  position  and  extension. 

7.  Calculus  (  Calculus,  a  pebble)  is  that   branch  of  mathe- 
matics which  commands  by  one    general   method,  the  most  diffi- 
cult problems  of  geometry  and  physics. 


12  FINKEL'S  SOLUTION  BOOK. 

8.  Differential   Calculus  is  that    branch  of  Calculus 
which  investigates   mathematical  questions  by  measuring  the  re- 
lation of  certain  infinitely  small  quantities  called  differentials. 

9.  Integral   Calculus  is  that  branch  of  Calculus  which 
determines  the  functions  from  which  a  given  differential  has  been 
derived. 

1C.  Calculus  of  Variations  is  that  branch  of  calculus 
in  which  the  laws  of  dependence  which  bind  the  variable  quanti- 
ties together  are  themselves  subject  to  change. 

11.  Quaternions    (quaternis,   from    quaterni    four   each, 
from  quator,  four)  is  that  branch  of  algebra  which  treats  of  the 
relations  of  magnitude  and  position  of  lines  or  bodies  in  space  by 
means  of  the  quotient  of  two  direct  lines  in  space,  considered  as 
depending  on  a  system  of  four  geometrical   elements,  and  as  ex- 
pressed by  an  algebraic  symbol  of  quadrinominal  form. 

12.  Platonic    Geometry  is  that  branch  of  geometry  in 
which  the  argument  is  carried  forward  by  a  direct  inspection  of 
the   figures  themselves,  delineated   before  the  eye,  or  held  in  the 
imagination. 

13.  Pure    Geometry  is  that  branch  of  Platonic  geometry 
in  which  the  argument  may  be  practically  tested  by  the  aid  of 
the  compass  and  the  square  only. 

14.  Conic   Sections  is  that  branch  of  Platonic  geometry 
which  treats  of  the  curved    lines   formed  by  the  intersection  of  a 
cone  and  a  plane. 

15.  Trigonometry      (rptytovov,       triangle,      ptrpov,     meas- 
ure) is  that  branch  of  Platonic  geometry  which  treats  of  the  re- 
lations of  the  angles  and  sides  of  triangles. 

16.  Plane  Trigonometry    is  that  branch  of  trigonom- 
etry which  treats  of  the  relations  of  the  angles  and  sides  of  plane 
triangles. 

17.  Analytical  Trigonometry  is  that  branch  of  trig- 
onometry  which  treats  of  the  general  properties  and  relations  of 
trigonometrical  functions. 

18.  Spherical  Trigonometry    is     that  branch  of  trig- 
onometry which  treats  of  the  solution  of  spherical  triangles. 

19.  Analytical    Geometry    is  that  branch  of  geometry 
in  which  the  properties  and  relations  of  lines  and  surfaces  are  in- 
vestigated by  the  aid  of  algebraic  analysis. 

20.  Descriptive    Geometry  is  that  branch  of  geometry 
which   seeks   the  graphic    solution  of   geometrical    problems  by 
means  of  projections  upon  auxiliary  planes. 


DEFINITIONS.  13 

21.  Mensuration  is  that  branch  of   applied  mathematics 
which  treats  of  the  measurment  of  geometrical'magnitudes. 

22.  Surveying    is    that    branch    of   applied     mathematics 
which  treats  of  the  art  of  determining  and  representing  distances, 
areas,  and  the  relative  position  of  points  upon  the  earth's  surface. 

23.  Navigation  is   that    branch  of  applied    mathematics 
which  treats  of   the  art  of   conducting  ships  from  one  place  to 
another. 

24:.      Mechanics    is    that   branch    of    applied    mathematics 
which  treats  of  the  laws  of  equilibrium  and  motion. 

25.  Astronomy       d.ffrpovop.ia^    from    affrpov,    star  and    VO/JLOS 
law)  is   that  branch  of  applied  mathematics  in  which    mechan- 
ical principles  are  used  to  explain  astronomical  facts. 

26.  Optics     (oTtTix-ij,     from     8</>ts    sight,)    is  that  branch  of 
applied   mathematics  which  treats  of  the  laws  of  light. 

27.  GrUnnery  is  that  branch  of  applied  mathematics  which 
treats  of  the  theory  of  projectiles. 

28.  A   Proposition  is  a  statement  of  something  proposed 
to  be  done. 

/       rm  \  !•  Lemma. 

,     T^  I,         (  a.  Ineorem.  \       on 

1.  Demonstrable.      \  ^  pro^jem     (  2-  Corollary. 

29.  Prop't'n.  - 

,  ,      (  a.  Axiom. 

2.  Indemonstrable.  <  ,  ^» 

/  £.  Postulate. 

30.  A   Demonstrable   Proposition  is    one   that   can 
be  proved  by  the  aid  of  reason. 

31.  A.   Theorem  is  a  truth  requiring  a  proof. 

32.  A   Lemma  is  a  theorem  demonstrated  for  the  purpose 
of  using  it  in  the  demonstration  of  another  theorem. 

33.  A    Corollary  is   a   subordinate    theorem,  the   truth  of 
which  is  made  evident  in   the  course  of  the   demonstration  of  a 
more  general  theorem. 

34.  A   Problem  is  a  question  proposed  for  solution. 

35.  An    Indemonstrable    Proposition  can   not  be 

proved  by  any  manner  of  reasoning. 

36.  An  Axiom  is  a  self-evident  truth. 

37.  A  Postulate  is  a  proposition  which  states  that  some- 
thing can  be  done,  and  which  is  so  evidently  true  as  to  require 
no  process  of  reasoning  to  show  that  it  is  possible  to  be  done. 


14  i-lNKEL'S  SOLUTION  BOOK. 

38.  A  Demonstration  is  the  process  of  reasoning,  prov- 
ing the  truth  of  a  proposition. 

39.  A   Solution  of  a  problem  is  an  expressed    statement 
showing  clearly  how  the  result  is  obtained. 

40.  An   Operation  is  a  process  of  finding,   from    given 
quantities,  others  that   are    known,  by  simply    illustrating    the 
solution. 

41.  A  Rule  is  a  general  direction  for  solving  all  problems 
of  a  particular  kind. 

42.  A  Formula  is   the  expression   of  a  general   rule  or 
principle  in  algebraic  language. 

43.  A   Scholium  is  a  remark  made  at  the   close  of  a  dis- 
cussion, and  designed  to  call  attention  to  some  particular  feature 
or  features  of  it. 


CHAPTER    II. 

NUMERATION    AND    NOTATION. 

1.     Numeration  is  the  art  of  reading  numbers. 

2.  There  are  two  methods  of  numeration  ;  the  French  and  the 
English. 

3.  The  French  method  is  that  in  general  use.    In  this  method, 
we  begin  at  the  right  hand  and  divide  the  number  into  periods 
of  three  figures  each,  and  give  a  distinct  name  to  each   period. 

4.  The  English  method  is  that  used  in  Great  Britain  and  the 
British  provinces..     In  this   method,  we  divide  the  number  (if  it 
consists  of  more  than  six  figures)  into  periods  of  six  figures  each, 
and  give  a  distinct  name  to  each  period.     The   following  number 
illustrates  the  two  methods  ;   the  upper  division  showing  how  the 
number  is  read  by  the  English  method,  and  the  lower  division 
showing  how  it  is  read  by  the  French  method. 

4th  period,     3d  period,     2d  period,     1st  period. 
Trillions.        Billions.        Millions.          Units. 

"^845  678^904    325^47   434^913 

5-3    5§     5g     -SS    ga     *.!    *p 
*~<§    ^o?     *        ^  ^EH    ^ 

5.  The  number  expressed  in  words  by  the  English  method, 
'eads  thus: 


NUMERATION  AND  NOTATION.  15 

Eight  hundred  forty-five  trillion,  six  hundred  seventy-eight 
thousand  nine  hundred  four  billion,  three  hundred  twenty-five 
thousand  one  hundred  forty-seven  million,  four  hundred  thirty- 
four  thousand  nine  hundred  thirteen. 

Remark. — Use  the  conjunction  and,  only  in  passing  over 
the  decimal  point.  It  is  incorrect  to  read  456,734  four  hundred 
and  fifty-six  thousand,  seven  hundred  and  thirty-four.  Omit  the 
and's&nd  the  number  will  be  correctly  expressed  in  words. 

6.     The  following  are  the  names  of  the  Periods,  according  to 
the  common,  or  French  method: 


First     Period,  Units. 

Second       "  Thousands. 

Third         "  Millions. 

Fourth       "  Billions. 

Fifth          "  Trillions, 


Sixth  Period,  Quadrillions. 
Seventh  "         Quintillions. 
Eighth    "         Sextillions. 
Ninth      "         Septillions. 
Tenth     "         Octillion. 


Other  periods  in  order  are,  Nonillions,  Decillions,  Undecil- 
lions,  Duodecillions,  Tredecilions,  Quatuordecillions  Quindecil- 
lions,  Sexdecillions,  Septendecillions,  Octodecillions,  Novende- 
cillions,  Vigintillions,  Primo-Vigintillions,  Secundo-vigintillions, 
Tertio-vigintillions,  Quarto- vigintillions,  Quinto-vigintillions, 
Sexto-vigintillions,  Septo-vigintillions,  Octo-vigintillions,  Nono- 
vigintillions,  Trigillions;  Primo-Trigillions,  Secundo-Trigillions, 
and  so  on  to  Quadragillions ;  Primo-quadragillions,  Secundo- 
quadragillions,  and  so  on  to  Quinquagillions;  Primo-quinqua- 
'gillions,  Secundo-quinquagillions,  and  so  on  to  Sexagillions, 
Pr  i  mo -sexagil  lions,  Secundo-sexagillions,  and  so  on  to  Septua- 
gillions  ;  Primo-septuagillions,  Secundo-septuagillions,  and  so  on 
to  Octogillions  ;  Primo-octogillions,  Secundo-octogillions,  and 
so  on  to  Nonogillions  ;  Primo-nonogillions,  Secundo-nonogillions, 
and  so  to  Centillions. 

7.     Notation  is  the  art  of  writing  numbers. 

There  are  three  methods  of  expressing  numbers ;  by  words, 
by  letters,  called  the  Roman  method,  and  by  figures,  called  the 
Arabic  method. 

8.  The  Roman  Notation,  so  called  from  its  having  originated 
with   the   ancient  Romans,  uses   seven  capital  letters  to  express 
numbers;  viz.,     I,  V,  X,  L,  C,  D,  M. 

9.  The  Arabic  Notation,  so  called  from  its  having  been  made 
known    through  the  Arabs,  uses  ten  characters  to  express  num- 
bers ;  viz.,     1,  2,  3,  4,  59  6,  7,  8,  9,  0. 

10.  Ordinal  Numbers.     A  logical  definition  of  number 
is  not  easy  to  give,  for  the  reason  that  the  idea  it  conveys  is  a 
simple  notion.     The  clearest  idea  of  what  counting  and  numbers 
mean  inay  be  gained  from  the  observation  of  children  and  of 


16  FINKEL'S  SOLUTION   BOOK. 

nations  in  the  childhood  of  civilization*.  When  children  count 
or  add  they  use  their  ringers,  or  small  sticks,  or  pebbles  which 
they  adjoin  singly  to  the  things  to  be  counted  or  otherwise  to  be 
ordinally  associated  with  them.  History  informs  us  that  the 
Greeks  and  Romans  employed  their  fingers  when  they  counted 
or  added.  The  reason  why  the  fingers  are  so  universally  used 
as  a  means  of  numeration  is,  that  everyone  possesses  a  definite 
number,  sufficiently  large  for  purposes  of  computation  and  that 
they  are  always  at  hand. 

Let  us  consider  the  row  of  objects,  XXXXXXXX 

XXXXXXXX ,  with  regard  to  their  order,  say 

from  left  to  right,  freeing  our  minds  from  all  notions  of  magni- 
tude. Beginning  with  any  one  object  in  this  row,  we  speak  of 
the  one  we  begin  with  as  being  the  first,  the  next  in  order  to  it 
to  the  right  the  second,  the  next  in  order  to  the  right  of  the  sec- 
ond the  third,  and  so  on.  The  name  or  mark  we  thus  attach  to 
an  object  to  tell  its  place  in  the  row  is  called  an  integer.  This 
process,  or  operation,  of  labeling  the  objects  is  called  counting 
and  it  is  the  fundamental  operation  of  mathematics. 
To  count  objects  is  to  label  the  objects,  not  primarily  to  tell  how 
many  there  aref.  In  thus  labeling  the  objects,  we  may  replace 
the  objects  by  the  fingers,  by  sticks,  by  pebbles,  by  marks,  or 
by  characters.  The  method  of  tallying  used  at  the  present  time 
is  such  a  method.  In  counting  objects  marks  are  made  until 
four  are  made,  then  these  are  crossed  with  a  fifth  mark  and  so 
on.  Thus  fH-F  -ffH  fH4. 

Suppose  that  in  counting  the  objects  in  the  row,  we  use  our 
fingers,  and  for  each  object  in  the  row  beginning  with  a  certain 
one  we  bring  in  correspondence  with  that  object  the  little  fin- 
ger of  the  right  hand,  with  the  next  object  to  the  right  the  next 
to  the  little  finger  of  the  right  hand,  and  so  on  until  an  object 
and  the  thumb  of  the  right  hand  are  brought  into  correspond- 
ence. For  the  group  of  objects  thus  counted,  let  us  bring  into 
correspondence  the  little  finger  of  the  left  hand.  Now  continue 
the  counting  of  the  objects  of  the  row  as  before,  and  when  a 
second  group  is  reached  bring  into  correspondence  with  this 
group  the  next  to  the  little  finger  of  the  left  hand.  Continue 
this  process  until  a  group  of  the  objects  as  represented  by  the 
fingers  of  the  right  hand  is  brought  into  correspondence  with 
the  thumb  of  the  left  hand.  Thus  the  fingers  of  the  left  hand 
represent  a  group  of  groups  of  objects.  Bring  this  group  rep- 
resented by  the  fingers  of  the  left  hand  into  correspondence  with 

*  Schubert's  Mathematical  Essays  and  Recreations. 

f  My  friend,  Dr.  William  Rullkoetter,  told  me  of  a  case  coming  under  his  personal 
observation,  where  a  farmer,  unable  to  count,  but  when  desirous  of  knowing:  if  any  of 
his  cattle  were  missing,  would  have  them  driven  through  a  gate  or  past  some  point 
where  he  could  see  them  as  they  passed  singly.  He  would  then  say, "  You  are  here," 
"  and  you  are  here,"  "  and  you  are  here,"  and  so  on  until  all  had  passed  by.  In  this  way 
he  was  able  to  tell  if  any  were  missing,  but  not  able  to  tell  how  many  he  had. 


NUMERATION   AND   NOTATION.  17 

the  little  toe  of  the  right  foot.  Now  continue  the  process  of 
counting  the  objects  and  so  on  as  before  until  the  big  toe  of  the 
right  foot  is  brought  into  correspondence  with  a  group  corre- 
sponding to  the  fingers  of  the  left  hand.  Thus  the  toes  of  the 
right  foot  represent  a  group  of  a  group  of  a  group  of  objects. 
In  this  manner,  we  could  build  up  the  system  of  numeration 
called  the  Quinary,  a  system  in  which  five  objects  as  represented 
by  the  fingers  of  the  right  hand  make  a  unit  or  group  as  repre- 
sented by  a  finger  of  the  left  hand,  five  groups  of  five  objects  as 
represented  by  the  fingers  of  the  left  hand  make  a  group  as  rep- 
resented by  the  toes  of  the  right  foot,  and  so  on. 

The  decimal  system  of  numeration  may  be  built  up  in  the 
same  way,  except  that  the  group  of  objects  corresponding  to  the 
fingers  of  both  hands  would  be  represented  by  a  toe.  After  the 
fingers  and  toes  have  been  exhausted  in  the  process  of  counting 
the  numeration  would  have  to  be  continued  by  using  small  sticks 
or  pebbles.  It  is  very  probably  due  to  the  fact  that  we  have  10 
fingers  .that  the  decimal  system  was  invented.  There  are,  how- 
ever, among  the  uncivilized  nations  of  the  world  a  number  of 
different  systems  of  numeration*. 

At  the  present  time,  in  labeling  objects  by  the  process  of 
counting  we  use  the  following  characters,  viz.,  1,  2,  3,  4,  5, 
6,  7,  8,  9,  etc. 

123456789 labels. 

Thus  XXXXXXXXXXXXXXX objects. 

In  labeling,  we  could  begin  .with  the  object  marked  3  and 
re-label  it  1,  then  re-label  4  as  2  and  5  as  3,  and  so  on.  This  is 
expressed  by  writing 

3—2=1,  4—2=2,  5—2=3, 

meaning  that  if  we  begin  after  the  object  whose  old  mark  was  2, 
then  the  object  which  was  third  becomes  first,  the  object  which 
was  fourth  becomes  second,  and  so  on.  Beginning  after  an 
object  instead  of  with  it  suggests  that  our  original  row  might 
begin  after  an  object;  this  object  after  which  the  counting  begins 
is  marked  0  and  called  the  origin.  If  there  are  objects  to  the 
left  of  the  origin,  we  count  them  in  the  same  way;  except  that 
we  prefix  the  sign,  — ,  to  show  that  they  are  to  the  left,  and  we 
call  the  marks  so  changed  negative  integers,  thus  distinguishing 
them  from  the  old  marks  which  we  call  positive  integers.  The 
marks  are  ....  —4,  —3,  —2,  --1,  0,  1,  2,  3,  4,  5,  6, 

These  marks  constitute  what  is  called  the  natural  integer- 
system. 

When  an  object  marked  a  is  to  the  left  of  another  marked 
a',  we  say  that  a  comes  before  a'  or  is  inferior  to  a',  and  a' 

*See  Conant's  Number  Concept  for  a  full   treatment  of  the  various  systems  of 
notation. 


18  FINKEL'S   SOLUTION   BOOK. 

comes  after  a  or  is  superior  to  a.  These  ideas  are  expressed 
symbolically  thus  a<a! ',  a>a.  Here  a  and  a'  mean  integers, 
positive  or  negative. 

Objects  considered  as  a  succession  from  left  to  right  are  in 
positive  order;  when  considered  from  right  to  left,  in  negative 
order. 

Addition  and  its  inverse  operation,  Subtraction,  are  algorithms 
of  counting.  Multiplication  is  an  algorithm  of  Addition,  and 
Division  is  an  algorithm  of  Subtraction.  Addition,  Subtraction, 
Multiplication,  and  Division  are  only  short  methods  of  counting. 

If  we  operate  on  any  integer  of  the  natural-integer  series  by 
any  one  of  the  operations  of  Addition,  Subtraction,  or  Multipli- 
cation, no  new  integer  is  produced.  With  reference  to  these 
operations  the  natural  integer-series  is  closed,  that  is  to  say, 
there  are  no  breaks  in  the  integer-series  into  which  other  inte- 
gers arising  from  these  operations  may  be  inserted.  If,  how- 
ever, we  operate  on  any  one  of  the  integers  of  the  integer-series 
by  the  operation  of  Division,  the  operations  in  many  cases  are 
impossible.  Suppose  we  wish  to  divide  17  by  5.  This  opera- 
tion is  absolutely  impossible.  *£-  is  a  meaningless  symbol  with 
reference  to  the  fundamental  operation  of  mathematics.  But  in 
this  case,  as  in  the  case  when  negative  numbers  are  introduced 
by  the  inverse  operation,  subtraction,  we  apply  a  principle  called 
by  Hankel,  "The  Principle  of  the  Permanence  of  Formal  Laws," 
and  by  Schubert,  "The  Principle  of  No  Exception,"  viz.,  That 
every  time  a  newly  introduced  concept  depends  upon  operations 
previously  employed,  the  propositions  holding  for  these  operations 
are  assumed  to  be  valid  still  when  they  are  applied  to  the  new  con- 
cepts. In  accordance  with  this  principle,  we  invest  the  symbol, 
-1/-,  which  has  the  form  of  a  quotient  without  its  dividend  being 
the  product  of  the  divisor  and  any  number  yet  defined,  with  a 
meaning  such  that  we  shall  be  able  to  reckon  with  such  apparent 
quotient  as  with  ordinary  quotients.  This  is  done  by  agreeing 
always  to  put  the  product  of  such  a  quotient  form  with  its 
divisor  equal  to  its  dividend.  Thus,  (^-)Xb=Vj.  We  thus 
reach  the  definition  of  fractions.  The  concept  of  fractions  may 
also  be  established  as  in  the  next  article. 

11.     Fractions.     Let   us   now  again   assume  the   row   of 

0123456789 

objects,  XXXXXXXXXXXX.  .   .  .  attending  to  only 

zero,  the  object  from  which  we  begin,  and  the  objects  on  the 
right  of  it.  Suppose  we  re-label  the  alternate  objects  2,  4,  6,  8, 
....  marking  them  1,2,8,4,  .  .  .  .  We  must  then  invent 
marks  for  the  objects  previously  marked  1,3,5,7 The 


NUMERATION   AND   NOTATION.  19 

marks  invented  are  shown  in  Figure  1,  above  the  objects,  the 
old  marks  being  below  the  objects. 

041f2  "|3}4J5 

xxxxxxxxxxxx.... 

0123456789 

Fig.  i. 

From  this  it  is  clear  that  instead  of  re-labeling  the  alternate 

objects  in  a  row  of  objects,  0,  1,  2,  3,  4,  5,  6,  7, we 

can  interpolate  alternate  objects  in  the  row  and  then  mark  them 
J,  f ,  f ,  and  so  on. 

In  the  same  way  we  can  interpolate  two  objects  between 

every  consecutive  two  of  the  given  row  0,  1,  2,  3,  4,  5, 

marking  the  new  objects  in  order  J,  §;  f,  |;  -J ,  -f ;   and  so  on. 

0  4  f  1  |  |  2  |  |  3 

Thus,        XXXXXXXXXX 

0123 

Fig.  2. 

-t  n          o 

In  this  way  we  account  for  the  symbols   —»—>—»    ...  where 

p    P .  P 

p  is  any  positive   integer.     These  we   call   positive   fractional 
numbers. 

By  interpolating  single  objects  in  the  row  0,  \,  1,  f ,  2,  f ,  .  .  . 
we  have  the  same  sequence  of  objects  as  if  we  interpolate  ob- 
jects by  threes  in  the  row 

0,  1,  2,  3,  4,  5,  6,  ...... 

and  the  objects  are  therefore  marked 

0,  i,  i,  i,  1, 

OJi|lffJ2 

Thus,        XXXXXXXXXX....^.. 

0        \        1        I        2 

Fig.  3. 

From  this  we  see  that  \  and  f  are  marks  for  the  same  object. 
Also  |  and  f .  Hence,  }=4  and  |=f . 

A  row  marked  0,  \,  f ,  J,  4,  f ,  1,  .  .  .  .  is  to  be  understood  as 
arising  from  the  interpolation  of  objects  by  fives;  that  is,  by 
introducing  the  objects  £,  f ,  f,  f,  f ,....,  or  £,  J,  \,  f ,  f,  .  .  .  . 

As  f  comes  before  f ,  we  say  f<f ,  or  J<i. 

We  may  interpolate  as  many  objects  as  we  please  in  the  nat- 
ural row,  and,  by  the  principle  of  the  least  common  denomina- 
tor, we  can  interpolate  so  as  to  explain  any  assigned  positive 
fractional  marks, /i,/2'/3»  •  •  •  Also,  given  any  positive  rational 
mark,  r,  other  than  zero,  we  can  interpolate  rational  marks  be- 


20  FINKEL'S   SOLUTION   BOOK. 

tween  0  and  r.  When  no  object  can  be  made  to  fall  between  an 
assigned  object  and  0,  that  assigned  object  must  be  0  itself. 

In  the  same  way  we  may  treat  the  negative  numbers. 

We  can  think  of  an  infinity  of  objects  as  being  interpolated 
in  the  natural  row,  so  that  each  shall  bear  a  distinct  rational 
number  and  so  that  we  can  say  which  of  any  two  objects  comes 
first.  It  is  to  be  noticed  that  as  we  approach  any  of  the  natural 
objects  there  is  no  last  fractional  mark;  that  is,  whatever  object 
we  take  there  are  always  others  between  it  and  the  natural  object. 

Thus,  if  an  infinitude  of  objects  be  interpolated  in  the  nat- 
ural row,  0,  1,  2,  3,'.  .  .  . 

0  *  i  HI 

XXXX . .  to  in finityXXX . .  to  infinityXXXXX . .  to  infinityXX . .  to  infinityX 
0  Fig.  4.  1 

then  it  is  clear  that  whatever  object  we  take  there  is  an  infinity 
of  objects  between  it  and  the  natural  object,  thus  rendering  it 
evident  that  there  are  no  last  fractional  marks  in  this  case. 

12.  Irrational  Numbers.  In  considering  square  num- 
bers from  the  ordinal  point  of  view,  we  re-label  our  natural  row 
as  in  Fig.  5. 

0   1   i        f        2  3 

xxxxxxxxxxxxxxxx 

0   1   i  2.  }   3  4    56789 

Fig.  5. 

where  the  old  names  are  below  and  the  new  above.  We  have 
now  to  consider  how  to  bring  the  omitted  objects  into  the  scheme 
of  ordinal  numbers.  Bvery  object  whose  new  name  is  fractional 
had  a  fractional  name,  so  that  the  object  whose  old  name  was  2 
cannot  now  have  a  rational  name.  We  give  it  a  name  which  we 
call  irrational.  _We  call  it  the  positive  or  chief  square  root  of  2 
and  mark  it  V  '2  or  2*.  As  an  ordinal  number  it  is  perfectly 
satisfactory,  for  we  know  where  it  comes,  whether  left  or  right 
of  any  proposed  rational  number,  by  means  of  the  old  marking. 
Hence,  it  separates  all  the  rational  numbers  into  two  classes, 
viz.,  those  on  its  right  and  those  on  its  left.  A  rational  number 
separates  all  other  rational  numbers  into  two  classes;  we  put  it 
into  one  of  the  classes  and  say  it  closes  that  class. 

Take,  for  example,  f .  Now,  there  is  no  last  fractional  mark 
as  we  approach  f  from  the  left  or  from  the  right.  Hence,  with- 
out %  neither  the  class  to  the  left  of  f  nor  the  class  to  the  right 
of  %  is  closed.  With  f ,  either  class  is  closed. 

Any  process  which  serves  to  separate  rational  numbers  into 
two  classes, — those  on  the  left  and  those  on  the  right,  such  that 
the  left-hand  class  is  not  closed  on  the  right  and  the  right-hand 
class  not  closed  on  the  left, — leads  to  the  introduction  of  a  new 
object  named  by  an  irrational  number. 


NUMERATION   AND   NOTATION.  21 

For  example,  V  2  separates  all  rational  numbers  into  two 
classes,  viz.,  those  on  the  left  of  it  and  those  on  the  right.  Now 
if  we  take  any  rational  object  however  near  to  the  V  2  as  we 
please  ^we  can  always  interpolate  new  rational  objects  between  it 
and  }/~2.  Thus,  it  is  clear  that  the  class  on  the  left  of  i/~2~is 
not  closed  at  the  right  nor  the  class  on  the  right  closed  on  the  left. 

Two  rational  or  irrational  numbers, — for  simplicity  take  them 
both  irrational  and  equal  to  s  and  /, — are  equal  if  the  rational 
objects  to  the  left  of  5-  are  the  same  as  the  rational  objects  to  the 
left  of  /,  and  the  rational  objects  to  the  right  of  s  are  the  same 
as  those  to  the  right  of  sr.  Thus,  4^  and  2^  effect  the  same  sep- 
aration of  the  rational  numbers.  Hence,  4^=2M. 

An  equivalent  condition  for  the  equality  of  ^  and  /  is  that 
every  rational  number  to  the  left  of  s  shall  be  to  the  left  of  /,  and 
every  rational  number  to  the  left  of  s'  shall  be  to  the  left  of  s. 

Between  two  unequal  irrational  objects,  s  and  /,  there  must 
lie  rational  objects;  for,  since  s  and  s'  are  not  equal,  there  must  be 
a  rational  number  which  is  before  one  and  not  before  the  other. 

It  is  very  important  to  notice  that  we  have  now  a  closed 
number-system.  When  we  seek  to  separate  the  irrational  objects 
as  lying  left  or  right  of  an  object,  either  the  object  is  rational  or 
if  not  it  separates  rational  objects  and  is  irrational;  in  any  case 
it  must  have  for  its  mark  a  rational  or  irrational  number,  and 
there  is  no  loop-hole  left  for  the  introduction  of  new  real  num- 
bers which  separate  existing  numbers.  This  is  often  briefly 
expressed  by  saying  that  the  whole  system  of  positive  and  neg- 
ative integral,  fractional,  and  irrational  numbers  is  continuous, 
or  is  a  continuum*. 

In  the  way  indicated  above,  the  number-concept  of  Arithme- 
tic is  put  on  a  basis  consistent  with  Geometry.  If  we  select  any 
point  on  a  straight  line  and  call  it  the  zero-point,  and  also  a  fixed 
length,  measured  on  this  line,  be  chosen  as  the  unit  of  length, 
any  real  number,  a,  can  be  represented  by  a  point  on  this  line  at 
a  distance  from  the  zero-point  equal  to  a  units  of  length.  Con- 
versely, each  point  on  the  line  is  at  a  distance  from  the  origin 
equal  to  a  units  of  length,  when  a  is  a  real  number.  That  is, 
there  is  a  one  to  one  correspondence  between  the  points  of  line 
and  the  numbers  of  the  real  number-system.  For  every  point 
of  the  line,  there  corresponds  a  number  of  the  real  number-sys- 
tem and  for  every  number  of  the  real  number-system  there  cor- 
responds a  point  of  the  line. 

EXAMPLES. 

1 .  Write  three  hundred  seventy  quadrillion,  one  hundred  one 
thousand  one  hundred  thirty-four  trillion,  seven  hundred  eighty- 

*See  Harkness  and  Morley's  Introduction  to  the  Theory  of  Functions,  Chapter  I., 
from  which  this  has  been  adapted. 


22  FINKEL  S  SOLUTION  BOOK 

nine  thousand  six  hundred  thirty-two  billion,  two  hundred  ninety- 
eight  thousand  seven  hundred  sixty-five  million,  four  hundred 
thirty-seven  thousand  one  hundred  fifty-six. 

2.  Read  by  the  English  method,  78943278102345789328903- 
24678. 

3.  Write  three  thousand  one   hundred  forty -one   quintillion, 
five  hundred  ninety-two  billion  six  hundred  fifty-three   million 
five  hundred  eighty-  nine  thousand  seven  hundred   ninety- three 
quadrillion,  two  hundred  thirty -eight  billion  four  hundred  sixty- 
two    million   six   hundred    forty-three    thousand    three    hundred 
eighty-three  trillion,  two  hundred  seventy-nine  billion  five  hun- 
dred two  million,  eight  hundred  eighty*-four  thousand  one  hundred 
ninety-seven. 

4.  Read  141421356237309504880168872420969807856971437- 
89132. 

5.  Is  a  billion,  a  million  million?     Explain. 

6.  Write  19  billion  billion  billion. 

7.  Write  19  trillion  billion  million  million. 

8.  Write  19  hundred  56  thousand. 

9.  Write  457  thousand  341  million. 

10.     Write  19  trillion  trillion  billion  billion  million  million. 


CHAPTEK    III. 

ADDITION. 

1.      Addition,  is  the  process  of  uniting  two  or  more  numbers 
of  the  same  kind  into  one  sum  or  amount. 

2.     Add  the  following,  beginning  at  the  right,  and  prove  the 
result  by  casting  out  the  9's  : 

7845  excess  of  9=6") 

6780   "  "  9=3 

8768   "  "  9=2  >Excess  of  9's=8. 

5343   "  "  9=6 

3987   "  "  9=OJ 

32723  excess  of  9=8 

Explanation. — Ad'ding  the  digits  in  the  first  number,  we 
have  24.  Dividing  by  9,  we  have  6  for  a  remainder,  which  is  the 
excess  of  the  9's.  Treating  the  remaining  numbers  in  the  same 
manner,  we  obtain  the  excesses  3,  2,  6,  0.  Adding  the  excesses 
and  taking  the  excess  of  their  sum,  we  have  8  ;  this  being  equal 
to  the  excess  of  the  sum  the  work  is  correct. 


3. 


SUBTRACTION. 

Add  the  following,  beginning  at  the  left : 

8456 
9799 
4363 
5809 
5432 


23 


33859 

From  this  operation,  we  see  that  it  is  more  convenient  to  be- 
gin at  the  right 

Remark. — We  can  not  add  8  apples  and  5  peaches  because 
we  can  not  express  the  result  in  either  denomination.  Only 
numbers  of  the  same  name  can  be  added. 

EXAMPLES. 

1.  Add    the    numbers    comprised    between    20980189    and 
20980197. 

2.  6095054  +  900703+90300420+9890655+37699+29753  = 
what? 

3.  Add   the  following,  beginning  at  the   left:     97674;   347- 
893;  789356;  98935679;  123456789. 

4.  Add  all  the  prime  numbers  between  1  and  107  inclusive. 

5.  Add  31989,  63060,  132991,  1280340, 987654321,  78903,  and 
prove  the  result  by  casting  out  the  9's. 

6.  Add  the  consecutive  numbers  from  100  to  130. 

7.  Add  the  numbers  from  9897  to  9910  inclusive. 

8.  Add   MDCCCLXXVI,   MDCXCVIII,    DCCCCXLIX, 
DCCCLXII. 


CHAPTER    IV. 

SUBTRACTION. 

1.      Subtraction  is  the  process  of  finding  the  difference  be- 
tween two  numbers. 

2.     Subtract   the    following    and    prove  the  result  by  casting 
out  the  9's  : 

984895  excess  of  9's=7 
795943      "       "  9's=ly 

188952       "       "  9'i 


,  ^Excess  of  9's=7. 
-6 


24  FINKEL'S  SOLUTION  BOOK. 

Explanation. — Adding  the  digits  in  the  first  number,  we 
have  43.  Dividing  by  9  the  remainder  is  7,  which  is  the  excess 
of  the  9's.  Treating  the  subtrahend  and  remainder  in  the  same 
manner,  we  have  the  excesses  1  and  6.  But  subtraction  is  the 
opposite  of  addition  and  since  the  minuend  is  equal  to  the  sum 
of  the  subtrahend  and  remainder,  the  excess  of  the  sum  of  the 
excesses  in  the  subtrahend  and  remainder  is  equal  to  the  excess 
in  the  minuend.  This  is  the  same  proof  as  that  required  if  we 
were  to  add  the  subtrahend  and  remainder. 

3.  We   begin  at  the  right  to  subtract,  so  that  if  a  figure  of 
the   subtrahend   is   greater   than   that   corresponding  to  it  in  the 
minuend,  we  can  borrow  one  from  the  next  higher  denomination 
and  reduce  it  to  the  required  denomination  and  then  subtract. 

4.  Subtract  the  following  and  illustrate  the  process  : 

1=9  99999   9+1        1=9  9999  9+1  )    A    ,  ,  1=9   9  9+1  )      .    , 

90000000    9856342  j-Add-   4326546  \  Add- 

85784895    8978567        3214957 


4215105     877775        11  11589 
EXAMPLES. 

1.  From  9347893987  take  8968935789.     Prove  the  result  by 
casting  out  the  9's. 

2.  7847893578— 6759984699— what  ? 

Which  is  the  nearer  number  to  920864;  1816090  or  27497? 

4.  34567—3451 8  +  3— -2  +  3—4  +  7  + 18—567  +  43812  — 1326  4 
678=what.  ? 

5.  5  +  6  +  7—12—13  + 14—2—3  +  7—8—6  +  5  + 12— 8— what  \ 

6.  3+4—  (6  +  7)—  8  +  27—  (1  +  3— 2— 3)  —  (7— 8  +  5)3  +  7=* 
what? 


CHAPTER    V. 

MULTIPLICATION. 

1.  Multiplication  is  the  process  of  taking  one  number  as 
many  times  as  there  are  units  in  another;  or  it  is  a  short  method 
of  addition  when  the  numbers  to  be  added  are  equal. 

2.     Multiply  the  following  and  prove   the  result  by  casting 
out  the  9's  : 

7855  excess  of  9's=7 
435       "        "  9's=3 

39275  21  excess  of  9's=3. 

23565 
31420 

3416925=excess  of  9*8=3. 


MULTIPLICATION.  25 

Explanation. — Adding  the  digits  in  the  multiplicand  and 
dividing  the  sum  by  9,  the  remainder  is  7  which  is  the  excess  of 
the  9's.  Adding  the  digits  in  the  multiplier  and  dividing  the  sum 
by  9,  we  have  the  remainder  3  which  is  the  excess  of  the  9's. 
Now,  since  multiplication  is  a  short  method  of  addition  when 
the  numbers  to  be  added  are  equal,  we  multiply  the  excess  in  the 
multiplicand  by  the  excess  in  the  multiplier  and  find  the  excess, 
and  this  being  equal  to  the  excess  in  the  product,  the  work  is 
correct. 

3.      Multiply  the  following,  beginning  at  the  left  : 

75645 
765 

1st.. 


2d 
3d 


57868425 

3.     From  this  operation,  we  see  that  it  is  more  convenient  to 
begin  at  the  right  to  multiply. 

5.  In  multiplication,  the  multiplicand  may  be  abstract,  or 
concrete;  but  the  multiplier  is  always  abstract. 

6.  The  sign  of  multiplication  is  X  >  and  is  read,  multiplied  by, 
or  times.     When  this  sign  is  placed  between  two  numbers  it  de- 
notes that  one  is  to  be  multiplied  by  the  other.       In  this  case,  it 
has  not  been   established  which  shall  be  the  multiplicand   and 
which  the  multiplier.     Thus  8x5=40,  either  may  be  considered 
the  multiplicand  and  the  other  the  multiplier.       If  8  is  the  mul- 
tiplicand, we   say,  8  multiplied  by  5  equals  40,  but  if  5  is  the 
multiplicand  we  say,  8  times  5  equals  40. 

EXAMPLES. 

1.  562402  X345728=what? 

2.  1  mile  =  63360  inches;   how  many  inches  from  the  earth 
to  the  moon  the  distance  being  239000  miles? 

3.  Multiply  789627  by  834,  beginning  at  the  left  to  multiply. 

4.  1  acre  =  43560  sq.  in.;  how  many  square  inches  in  a  field 
containing  427  acres? 


26  FINKEL'S  SOLUTION   BOOK. 

5.  Multiply  6934789643  by  34789.     Prove  the  result  by  cast- 
ing out  the  9's. 

6.  2778588  X  34678=what  ? 

7.  2X3X4— 3x7+3— 2X2+4+8X2+4— 3  X5+27=what? 

8.  5X7+6X7+8X7— 4X6+6  X  6+7  X6=what? 

9.  356789  X4876=what? 

10.  395076  X  576426=what  ? 

11.  7733447  X998800=what? 

12.  5654321X999880— what? 


CHAPTER    VI. 

DIVISION. 

1.  Division,  is  the  process  of  finding  how  many  times  one 
number  is  contained  in  another;  or,  it  is  a  short  method  of  sub- 
traction when  the  numbers  to  be  subtracted  are  equal. 

2.     Divide  the  following  and  prove  the  result  by  casting  out 
the  9's: 

67)5484888(81864 
536 

Dividend 

124  5484888  excess  of  9's=0. 

67 

Quotient 

578  81864  excess  of  9's=01  Excess   of    9's 

536  Un  this  product 

Divisor  i     ,-. 

428  67  excess  of  9's=4j   equals  0. 

402 


268 
268 

Explanation. — Adding  the  digits  in  the  dividend  and  di- 
viding the  sum  by  9,  we  have  the  remainder  0,  which  is  the  ex- 
cess of  the  9's.  Adding  the  digits  in  the  quotient  and  dividing 
the  sum  by  9,  we  have  the  remainder  0,  which  is  the  excess  of 
the  9's  in  the  quotient.  Adding  the  digits  in  the  divisor  and 
dividing  the  sum  by  9,  we  have  the  remainder  4,  which  is  the 
excess  of  the  9's  in  the  divisor.  Since  division  is  the  reverse  of 
multiplication,  the  quotient  corresponding  to  the  multiplicand, 
the  divisor  to  the  multiplier,  and  the  dividend  to  the  product,  we 
multiply  the  excess  in  the  quotient  by  the  excess  in  the  divisor. 
The  excess  of  this  product  is  0.  This  excess  being  equal  to  the 
excess  of  the  9's  in  the  dividend,  the  work  is  correct. 


DIVISION.  27 

If  there  be  a  remainder  after  dividing,  find  its  excess  and  add 
it  to  the  excess  of  the  product  of  the  excesses  of  the  quotient  and 
divisor.  Take  the  excess  of  the  sum  and  if  it  is  equal  to  the  ex- 
cess of  the  dividend  the  work  is  correct. 

3.  The  sign  of  division  is  -J-,  and  is  read  divided  by. 

4.  When  the  divisor  and  dividend  are  of  the  same  denomina- 
tion the  quotient  is  abstract ;  but  when  of  different  denomina- 
tions, the  divisor  is  abstract  and  the  quotient  is  the  same  as  the 
dividend.     Thus,  24  ct. -Met.  =  6,  and  24ct.-^4  =  6  ct. 

Remark. — We  begin  at  the  left  to  divide,  that  after  finding 
how  many  times  the  divisor  is  contained  in  the  fewest  left-hand 
figures  of  the  dividend,  if  there  be  a  remainder  we  can  reduce  it 
to  the  next  lower  denomination  and  find  how  many  times  the 
divisor  is  contained  in  it,  and  so  on. 

Note. — The  proof  by  casting  out  the  9's  will  not  rectify 
errors  caused  by  inserting  or  omitting  a  9  or  a  0,  or  by  interchang- 
ing digits. 

EXAMPLES. 

1.  4326422-f-961=what?     Prove   the   result   by  casting  out 
the  9's. 

2.  245379633477-r-1263=what?     Prove  the  result  by  casting 
out  the  9's. 

3.  What   number  multiplied  -by  109  with  98   added   to  the 
product,  will  give  106700? 

4.  The   product  of  two  numbers  is  212492745  ;  one  of  the 
numbers  is  1035;  what  is  the  other  number? 

5.  27-f-9  X  3-7-9  — 1+3-^3x9— 8-:-4+5X  2—3  X2-T-2---3-- 
(3X4-J-6+5— 2)+81-r-27x3-f-9Xl8«-6= what?      [Hint.—  Per- 
form the  operations  indicated  by  the  multiplication  and  division 
signs  in  the  exact  order  of  their  occurrence.] 

6.  (64-4-32X96-7-12  — 7—5+3)  Xj[(27-:-3)-f-9  — 1+2]  + 
93H-13X7— 45 }  X9+45-T-9+3— l=what?. 

7.  2x2-r-2-4-2-7-2x2X2-r-2-r-2-r-2=what?  Ans.  £. 

8.  3-r-3-f-3x3x3-7-3-r-OX4x4x5x5=what?  Ans.  oo. 

9.  2x2x2-f-2x2-7-2-r-2X2X2XOX2X2=what?      Ans.   0. 


28 


FINKEL'S  SOLUTION  BOOK. 


CHAPTER    VII. 

COMPOUND  NUMBERS. 

1.  A   Compound  Number  is  a  number  which  expresses 
several  different  units  of  the  same  kind  of  quantity. 

2.  A    Denominate    Number    is  a  concrete  number   in 
which  the  unit  is  a  measure;  as,  5  feet^  7  pints. 

3.  The    Terms  of  a  compound  number  are  the  numbers  of 
its  different  units.      Thus,  in  4  bu.    3  pk.    7  qt.    1  pt,  the  terms 
are  4  bu.  and  3  pk.  and  7  qt.  and  1  pt. 

4.  Reduction  of  Compound  Numbers  is  the  process 
of  changing  a  compound  number  from  one  denomination  to  an- 
other.    There    are  Two  Cases,  Reduction  Descending"  and  Re- 
duction Ascending. 

5.  Reduction  Descending  is  the  process  of  reducing  a 
number  from  a  higher  to  a  lower  denomination. 

6.  Reduction  -Ascending  is  the   process  of  reducing  a 
number  from  a  lower  to  a  higher  denomination. 

Ex.     Reduce  2  E.  Fr.  1  E.  En.  2  E.  Fl.  3  yd.  2  na.  to  nails. 


E.  Fr.  E.  En.  E.  Sc.  E.  Fl.  yd.  qr.  na. 
21  232 

65  34 


12qr.       5qr. 


6qr.  12  qr. 
6" 
5" 


35  qr. 
4 

140  na. 

2" 

I42na. 


TIME    MEASURE.  29 


TIME   MEASURE. 

I.  Time  is  a  measured  portion  of  duration. 

2.     The  measures  of  time  are  fixed  by  the  rotation  of  the  earth 
on  its  axis  and  its  revolution  around  the  sun. 

3.  A  Day  is  the  time  of  one  rotation  of  the  earth  on  its 
axis. 

4.  A.    Year  is    the    time    of   one    revolution    of    the   earth 
around  the  sun. 

TABLE. 

60  seconds  (sec.)  make  1  minute  (min.) 

60  minutes  "  1  hour  (hr.) 

24  hours  "  1  day  (da.) 

7  days  "  1  week  (wk.) 

4  weeks  "  1  lunar  month  (mo.) 

13  lunar  months,  1  da.  6  hr.  "  1  year  (yr.) 

12  calender  months  •'  1  year. 

365  days  "  1  common  year. 

365  da'.  5  hr.  48  min.  46.05  sec.    "  1  solar  year. 

365  da.  6  hr.  9  min.  9  sec.  "  1  sidereal  year. 

365  da.  6  hr.  13  min.  45.6  sec.  "  1  Anomalistic  year. 

366  days  "  1  leap  year,  or  bissextile  year. 
354  days                                      ,  "  1  lunar  year. 

19  years  "  1  Metonic  cycle. 

28  years  "  1  solar  cycle. 

15  years  "  1  Cycle  of  Indiction. 

532  years  1  Dionysian  Period. 

5.  The  unit  of  time  is  the  day. 

6.  The  Sidereal  Day  is  the  exact  time  of  one   rotation 
of   the  earth  on  its  axis.        It  equals  23  hr.  56  min.  4.09  sec. 

7.  The  Solar  Day  is  the  time  between  two  successive 
appearances  of  the  sun  on  a  given  meridian. 

8.  The  Astronomical  Day  is   the   solar    day,  begin- 
ning and  ending  at  noon. 

9.  The  Civil  Day9  or  Mean  Solar  Day9  is  the  average 
of  all    the    solar    days    of  the    year.      It  equals  24    hr.  3   min. 
56.556  sec. 

1O.  The  Solar  Year,  or  Tropical  Year9  is  the  time 
between  two  successive  passages  of  the  sun  through  the  vernal 
equinox. 

II.  The  Sidereal  Year  is  the  time  of  a  complete  revolu- 
tion of  the  earth  about  the  sun,  measured  by  a  fixed  star. 

13.  The  Anomalistic  Year  is  the  time  of  two  suc- 
cessive passages  of  the  earth  through  its  perihelion. 

13.  A  Lunar  Year  is  12  lunar  months  and  consists  of 
354  day. 


30  FINKEL'S  SOLUTION  BOOK. 

14.  A  Metonic   Cycle  is  a  period  of  19  solar  years,  after 
which  the  new  moons  again  happen  on  the  same  days  of  the  year. 

15.  A  Solar  Cycle  is  a  period  of  28  solar   years,  after 
which  the   first    day    of  the  year  is   restored   to   the   same   day 
of  the  week.      To  find   the  year  of  the   cycle,  we  have  the  fol- 
lowing rule: 

Add  nine  to  the  date,  divide  the  sum  by  twenty -eight;  the 
quotient  is  the  number  of  cycles,  and  the  remainder  is  the  year  of 
the  cycle.  Should  there  be  no  remainder  the  proposed  year  is  the 
twenty-eighth,  or  last  of  the  cycle.  The  formula  for  the  above 

rule  is  <!  \   in    which    x    denotes    the    date,   and   r    the    re- 

mainder which   arises   by  dividing   x-\-§  by  28,  is   the   number 
required. 

Thus,  for  1892,  we  have  (1 892+9 )-^28=67ff  •'•  1892  is  the 
25th  year  of  the  68  cycle. 

16.  The  Lunar  Cycle  is  a  period  of  19  years,  after  which 
the  new  moons  are  restored  to  the  same  day  of  the  civil  month. 

The  new  moon  will  fall  on  the  same  days  in  any  two  years 
which  occupy  the  same  place  in  the  cycle;  hence,  a  table  of  the 
moon's  phases  for  19  years  will  serve  for  any  year  whatever 
when  we  know  its  number  in  the  cycle.  This  number  is  called 
the  Golden  Number. 

To  find  the  Golden  Number  :  Add  1  to  the  date,  divide  the 
sum  by  19;  the  quotient  is  the  number  of  the  cycle  elapsed  and  the 
remainder  is  the  Golden  Number. 

r*+ii 

The  formula  for  the  same  is  <j  \    in   which  r  is  the   re- 

mainder after   dividing   the   date-}-l   by   19.     It    is   the   Golden 
Number. 

17.  A  Dionysian  or  Paschal  Period  is  a  period  of 
532  year,  after  which  the  new  moons  again  occur  on  the  same 
day  of  the  month  and  the  same  day  of  the  week.     It  is  obtained 
by  multiplying  a  Lunar  Cycle  by  a  Solar  Cycle. 

18.  A   Cycle  Of  Induction  is  a  period  of   15  years,  at 
the  end  of  which  certain  judicial  acts  took  place  under  the  Greek 
emperors. 

19.  JEpact  is  a  word  employed   in  the  calender  to  signify 
the  moon's  age  at  the  beginning  of  the  year. 

The  common  solar  year,  containing  365  days,  and  the  lunar 
year  only  354,  the  difference  is  11  days;  whence,  if  a  new 'moon 
fall  on  the  first  of  January  in  any  year,  the  moon  will  be  11  days 
old  on  the  first  day  of  the  following  year,  and  22  days  on  the 
first  of  the  third  year.  The  epact  of  these  years  are,  therefore, 
eleven  and  twenty-two  respectively.  Another  addition  of  eleven 


LONGITUDE  AND  TIME, 


31 


lays  would  give  thirty-three  for  the  epact  of  the  fourth  year;  but 
in  consequence  of  the  insertion  of  the  intercalary  month  in  each 
third  year  of  the  lunar  cycle,  this  epact  is  reduced  to  three.  In 
like  manner  the  epacts  of  all  the  following  years  of  the  cycle  are 
obtained  by  successively  adding  eleven  to  the  epact  of  the  former 
year,  and  rejecting  thirty  as  often  as  the  sum  exceeds  that 
number. 

LONGITUDE  AND  TIME. 

In  the  diagram,  the  curve  ACBD  represents  the  path  of  the 
earth  in  its  journey  around  the  sun.  This  curve  is  called  an 
ellipse.  The  ellipse  may  be  drawn  by  taking  any  two  points 
.Fand  F'  and  fastening  in  them  the  extremities  of  a  thread  whose 


length  is  greater  than  the  distance  F' F.  Place  the  point  of  a 
pencil  against  the  thread  and  slide  it  so  as  to  keep  the  thread 
constantly  stretched ;  the  point  of  the  pencil  in  its  motion  will 
describe  the  ellipse. 

The  points  F  and  F'  are  called  the/0«,  the  plural  of  focus. 

The  sun  occupies  one  of  these  foci.  The  plane  of  the  earth's 
orbit,  or  path,  is  called  the  ecliptic.  When  the  earth  is  at  A  it  is 
nearer  the  sun  than  when  it  is  at  B.  When  the  earth  is  nearest 
the  sun  it  is  said  to  be  in  perihelion  (Gr.  Kepi=peri,  near,  and 
fytos=halios,  sun);  when  farthest  from  the  sun,  it  is  said  to  be 
in  aphelion  (Gr.  dn6=apo,  from,  and  r\kw~-=halios,  sun).  The 
points  A  and  B,  in  the  diagram,  represent  the  perihelion  and 
aphelion  distances,  respectively.  The  earth  is  nearest  the  sun 
about  the  first  of  January  and  farthest  from  the  sun  about  the 
first  of  July.  It  takes  the  earth  365  da.  6  hr.  13  min.  45.6  sec.  to 
travel  from  A,  west  around  through  C,  B,  and  D  back  to  A. 


32  FINKEL'S  SOLUTION  BOOK. 

This  period  of  time  is  called  the  anomalistic  year.  The  west 
point  as  here  spoken  of,  may  be  thought  of  thus :  Suppose  you 
were  located  at  some  point  on  the  surface  of  the  sun  in  a  posi- 
tion enabling  you  to  see  the  North  Star.  Then  if  you  should 
face  that  star  you  would  be  facing  north,  your  right  hand  would 
be  to  the  east,  and  your  left  hand  to  the  west,  and  south  to 
your  back. 

While  the  earth  makes  one  revolution  around  the  sun,  it 
rotates  366  times  on  one  of  its  diameters.  The  diameter  upon 
which  it  rotates  is  called  its  axis.  The  axis  of  the  earth  is  in- 
clined from  a  perpendicular  to  the  plane  of  the  earth's  orbit  at 
an  angle  of  28J°.  If  the  axis  of  the  earth  were  extended  indefi- 
nitely, it  would  pass  very  near,  1J°,  from  the  North  Star. 

The  earth's  axis  and  the  sun  determine  a  plane,  and  this 
plane  is  of  great  importance  in  gaining  a  thorough  understand- 
ing of  Longitude  and  Time.  Suppose  you  are  on  the  earth's 
surface,  facing  the  sun  and  in  this  plane.  Then  you  will  have 
noon,  while  just  to  the  east  of  you  it  will  be  after  noon  and  just 
to  the  west  it  will  be  before  noon.  The  intersection  of  this  plane 
with  the  earth's  surface  is  called  the  trace  of  the  plane  on  the 
earth's  surface.  This  trace  is  called  a  meridian. 

If  you  could  travel  in  such  a  way  as  to  remain  in  this  plane 
for  a  whole  day,  that  is,  24  hours,  you  would  have  noon  during 
the  whole  time.  But  if  you  remain  stationary  on  the  earth's 
surface,  you  will  be  carried  out  of  this  plane  eastward  by  the 
earth's  rotation.  You  may  conceive  that  you  are  at  Greenwich, 
formerly  a  small  suburban  town  of  London,  Eng.,  but  now  in- 
corporated in  that  city,  with  the  sun  visible  and  having  such  a 
position  that  you  are  in  the  plane  formed  by  the  earth's  axis  and 
the  sun.  It  will  then  be  noon  to  you  at  that  place. 

Suppose  we  take  the  trace  of  the  plane,  in  this  position,  on 
the  earth's  surface  as  our  standard  meridian.  Then  all  places 
east  of  this  line  will  have  had  noon  and  all  places  west  of  it 
are  yet  to  have  noon.  As  the  earth  continues  to  rotate,  rotating 
as  it  does  from  west  to  east,  it  will  bring  points  west  of  the 
plane  into  coincidence  with  the  plane  and  thus  these  points  will 
have  noon  successively  as  they  come  into  the  plane.  Suppose 
we  start  when  Greenwich  is  in  this  plane,  and  mark  the  trace  of 
the  plane  on  the  earth's  surface  and  every  four  minutes  we  mark 
the  trace  of  the  plane;  in  this  way,  in  a  complete  rotation  of  the 
earth,  we  will  have  drawn  360  of  these  traces,  which  we  have 
agreed  to  call  meridians. 

The  distance  of  these  lines  apart,  measured  on  the  equator, 
is  called  a  degree  of  longitude,  better  an  arc-degree  of  longitude. 
Instead  of  measuring  longitude  from  Greenwich  entirely  around 
the  earth  through  the  west,  we  generally  measure  it  east  and 
west  to  180°. 


LONGITUDE  AND  TIME.  33 

Thus,  a  place  located  on  the  70th  meridian,  west,  is  said  to 
be  70°  west  longitude,  and  a  place  situated  on  the  195th  merid- 
ian, counting  from  Greenwich  around  through  the  west,  is  said 
to  be  85°  east  longitude. 

From  the  above  discussion,  we  see  that,  since  the  earth  turns 
on  its  axis  once  in  24  hours, 

24  hrs.  =360°  of  long.,  or  360°  of  long.=24  hrs. 
1  hr.    =^3-  of  360°=! 5°  of  long.,  or  15°  of  long.=l  hr. 
1  min.=gJ0-  of    15°=15'   of  long.,  or  15'  of  long.=l  min. 
1  sec.  =fa  of    15' =15"  of  long.,  or  15"  of  long.=l  sec. 

Hence,  if  we  have  the  difference  of  longitude  of  two  places,  we 
can  readily  find  the  difference  of  time  between  these  two  places. 

For  example,  the  longitude  of  St.  Petersburg  is  30°  16'  E., 
and  the  longitude  of  Washington  is  77°  0'  36"  W.  Now  the  dif- 
ference of  longitude  between  these  two  places  is  77°  0'  36"  + 
30°  l&=m°  16'  36".  Hence,  since  15°=1  hour,  107°  16'  36"= 
(107°  16'  36")-KL5f  or  7  hrs.  9  min.  6.4  sec.,  which  is  the  differ- 
ence of  time  between  Washington  and  St.  Petersburg. 

Conversely,  if  we  know  the  difference  of  time  between  two 
places,  we  can  easily  find  the  difference  of  longitude. 

For  example,  the  difference  of  time  between  New  York  City 
and  St.  Louis  is  1  hr.  4  min.  47J  sec.  Find  the  difference  of 
longitude. 

1  hr.     =15°. 
1  min.  =15'. 
II.  1  3.     4  min.  =60',  or  1°. 

Isec.    =15". 
5.   47J-  sec.  =47JX15"=710"=11'  50". 

III.      Hence,  the  difference  of  longitude  is  16°  11'  50". 

In  some  cases,  problems  are  so  proposed  that  we  are  to  find 
the  longitude  or  time  of  one  place,  having  given  the  longitude 
or  time  of  another  place  and  the  difference  of  time  or  difference 
of  longitude  of  the  two  places.  Such  problems  require  no  prin- 
ciples beyond  those  already  established. 

For  example,  the  difference  of  time  between  two  places  is 
2  hr.  30  min.  The  longitude  of  the  eastern  place  is  56°  W. 
Find  the  longitude  of  the  western  place. 

1.  Ihr.    =15°. 

2.  2hr.    =30°. 

3.  1  min.=15'. 

II.  <!  4.    30  min.=30X15'=450=7°  30'. 

5.  .'.  37°  30'=difference  of  longitude. 

6.  .'.  56°+37°  30'=93°  30',  the  longitude  of  the  west- 

ern place. 


34  FINKEL'S  SOLUTION  BOOK. 

Had  the  place  whose  longitude  is  given  been  in  east  longi- 
tude, we  would  have  subtracted  the  difference  of  longitude  to 
find  the  longitude  of  the  western  place. 

The  following  suggestions  may  prove  helpful  in  the  solution 
of  problems  in  Longitude  and  Time : 

1.  When  the  longitude  of  a  place  is  required,  having  given 
the  longitude  of  some  other  place  and  the  difference  of  longitude 
between  the  two  places. 

Conceive  yourself  located  at  the  place  whose  longitude  is 
given.  Then  ask  yourself  this  question,  Is  the  place  whose 
longitude  is  required,  east  or  west?  If  west,  add  the  difference 
of  longitude  when  the  given  longitude  is  west,  and  subtract  if 
the  given  longitude  is  east.  If  the  answer  to  your  question  is 
east,  subtract  the  difference  of  longitude  when  the  given  longi- 
tude is  west  and  add  the  difference  of  longitude  when  the  given 
longitude  is  east. 

If  the  places  are  on  opposite  sides  of  the  standard  meridian, 
subtract  the  given  longitude  from  the  difference  of  longitude  and 
the  difference  will  be  the  longitude  required,  and  will  be  oppo- 
site in  name  from  the  given  longitude.  That  is  to  say,  if  the 
given  longitude  is  east,  the  required  will  be  west,  and  vice  versa. 

2.  When  the  time  of  place  is  required,  having  given  the  time  at 
some  other  place  and  the  difference  of  time  between  the  two  places. 

Conceive  yourself  located  at  the  place  whose  time  is  given. 
Then  ask  yourself  this  question,  Is  the  place  whose  time  is 
required,  east  or  west  of  me?  If  the  answer  to  your  question 
is  west,  subtract  the  difference  of  time  for  the  required  time. 
If  the  answer  to  your  question  is  east,  add  the  difference  of 
time  for  the  required  time. 

STANDARD    TIME. 

In  1883,  the  railroad  officials  of  the  United  States  and  Canada 
adopted  what  is  called  standard  time.  These  officials  agreed  to 
adopt  the  solar  time  of  some  standard  meridian  as  the  local  time 
of  an  extended  area.  The  standard  meridians  thus  adopted  are 
75th,  90th,  105th,  and  120th.  All  stations  in  the  belt  of  country 
7^-°  wide  on  either  side  of  these  standard  meridians  have  as  local 
time  the  solar  time  of  the  respective  meridian.  For  example, 
all  points  or  stations  in  the  belt  of  country  7J°  wide  on  either  side 
of  90th  meridian,  i.  e.,  the  belt  of  country  lying  between  82^° 
and  97-|°  west  longitude  have  as  local  time  the  solar  time,  or  sun 
time,  of  the  90th  meridian.  In  other  words,  all  time-pieces  of 
the  various  stations  in  this  belt  indicate  the  same  time  of  day  as 
clocks  in  the  depots  situated  on  the  90th  meridian.  For  exam- 
ple, when  the  clock  m  the  Union  Depot  at  St.  Louis  indicates 
noon,  12  o'clock  M.,  the  clocks  in  the  Union  Depots  at  Indian- 


STANDARD  TIME.  35 

apolis  and  Kansas  City  also  indicate  noon,  though  at  Indianap- 
olis it  is  a  little  more  than  16  min.  past  noon  and  at  Kansas  City 
it  is  a  little  more  than  17  min.  till  noon,  sun  time.  That  is,  the 
standard  time  and  local  time  at  Indianapolis  differ  by  a  little 
more  than  16  min.,  standard  time  being  about  16  min.  slower 
than  local  time,  and  at  Kansas  City  standard  time  and  local  time 
differ  by  about  17  miu.,  standard  time  being  about  17  min.  faster 
than  local  time. 

If  one  were  to  set  his  watch  with  the  railroad  clock  in  the 
depot  at  Columbus,  Ohio,  then  take  the  train  for  Springfield, 
Mo.,  on  arriving  at  Springfield,  Mo.,  one  would  find  that  his 
watch  agrees  with  the  clock  in  the  Frisco  depot.  This  is  be- 
cause Columbus,  Ohio,  and  Springfield,  Mo.,  are  located  in  the 
belt  of  country  having  central  time,  i.  e.,  having  the  sun  time  of 
the  90th  meridian. 

How  about  the  local  time  of  these  two  places?  The  local 
time  at  Columbus,  O.,  is  about  28  min.  faster  than  standard  time. 
The  sun  comes  to  the  meridian  of  Columbus  before  it  comes  to 
the  90th  meridian.  When  the  sun  comes  to  the  meridian  at 
Columbus  it  is  noon,  local  time,  but  it  will  not  be  noon,  standard 
time,  until  the  sun  comes  to  the  90th  meridian,  which  will  be 
about  28  min.  later.  Hence,  local  time  at  Columbus,  O. ,  is  about 
28  min.  faster  than  standard  time.  A  passenger  going  from 
Columbus,  Ohio,  to  Springfield,  Mo.,  and  carrying  standard  time 
of  Columbus,  would  have  standard  time  at  St.  L,ouis,  standard 
time  and  local  time  at  St.  L<ouis  being  very  nearly  the  same.  On 
arriving  at  Springfield,  Mo.,  his  watch  would  still  indicate  stand- 
ard time  and  would  agree  with  the  regulator  in  the  depot  at 
Springfield,  but  his  time  would  be  about  8  min.  faster  than  the 
local  time  at  Springfield. 

The  time  in  the  belt  of  country  between  67J-0  west  longitude 
and  82^°  west  longitude  is  called  Eastern  Time ;  between  82-^-° 
W.  and  97i°  W.,  Central  Time;  between  97^-°  W.  and  112£°  W., 
Mountain  Time;  and  between  11 2^°  W.  and  127J°  W.,  Pacific 
Time.  We  might  call  the  time  in  the  belt  between  7^°  E.  and 
7£°  W,.  Greenwich  Time;  between  7J°  W.  and  22|°  W.,  East 
Atlantic  Time;  between  22^°  W.  and  37^°  W.,  Central  Atlantic 
Time;  between  37J°  W.  and  52^°  W.,  West  Atlantic  Time;  and 
between  52^°  W.  and  67JC  W.,  ^Colonial  Time. 

By  some  appropriate  system  of  nomenclature,  the  naming  of 
the  time  in  the  belt  beginning  with  127^°  W.  longitude  might 
be  extended.  However,  these  names  would  have  a  very  limited 
use  and  are  therefore  not  worth  coining. 


36  FINKEIv'S  SOLUTION  BOOK. 


THE   INTERNATIONAL  DATE   LINE. 

The  International  Date  Line  is  an  irregular  line  pass- 
ing through  Bering  Strait,  along  the  coast  of  Asia  to  near 
Borneo  and  Philippine  Islands,  and  thence  along  the  northern 
limits  of  the  East  Indian  Islands,  New  Zealand,  and  New  Guinea. 
It  is  the  line  from  which  every  date  on  the  earth  is  reckoned. 
At  present,  however,  the  180th  meridian  is  very  generally  used 
in  its  stead. 

Suppose  one  were  standing  on  the  180th  meridian  at  the  time  it  is 
noon,  Wednesday  (say),  at  Greenwich,  and  facing  north.  Then,  just  to 
right,  or  east  of  this  line,  Wednesday  is  beginning,  i.  e.,  Wednesday  12 
o'clock,  A.  MM  while  just  to  left,  or  west  of  the  line,  Wednesday  is  ending, 
i.  e.,  Wednesday  12  o'clock,  P.  M.  The  difference  of  time  between  places 
immediately  east  and  immediately  west  of  the  line  is  therefore  24  hours, 
and  just  west  of  the  line  it  is  one  day  later  than  just  east  of  it.  This  is 
made  still  clearer  by  considering  what  takes  place  as  the  earth  rotates  on 
its  axis;  the  places  just  west  of  the  line  will  be  carried  eastward,  and  since 
these  places  had  Wednesday  ending,  they  must  now  have  Thursday  begin- 
ning. But  these  places  are  west  of  the  line,  the  places  east  of  the  line 
still  having  Wednesday.  Hence,  it  is  clear  that  it  is  one  day  later  just 
west  of  the  line  than  just  east  of  the  line.  In  crossing  this  line,  there- 
fore, from  the  east  one  day  must  be  added,  while  in  crossing  it  from  the 
west  one  day  must  be  subtracted. 

Professor  C.  A.  Young,  in  his  General  Astronomy,  in  answering  the 
question,  "Where  does  the  day  begin?"  says,  "If  we  imagine  a  traveler 
starting  from  Greenwich  on  Monday  noon  and  traveling  westward  as  swiftly 
as  the  earth  turns  to  the  east  under  his  feet,  he  would,  of  course,  keep  the 
sun  exactly  on  the  meridian  all  day  long  and  have  continual  noon.  But 
what  noon  ?  It  was  Monday  when  he  started,  and  when  he  gets  back  to 
Ivondon,  twenty-four  hours  later,  it  is  Tuesday  noon  there,  and  there  has 
been  no  intervening  sunset.  When  does  Monday  noon  become  Tuesday 
noon?  The  convention  is  that  the  change  of  date  occurs  at  the  180tb 
meridian  from  Greenwich.  A  ship  crossing  this  line  from  the  east  skips 
one  day  in  so  doing.  If  it  is  Monday  forenoon  when  the  ship  reaches  the 
line,  it  becomes  Tuesday  forenoon  the  moment  it  passes  it,  the  intervening 
twenty-four  hours  being  dropped  from  the  reckoning  on  the  log-book. 
Vice  versa,  when  a  vessel  crosses  the  line  from  the  western  side,  it  counts 
the  same  day  twice,  passing  from  Tuesday  forenoon  back  to  Monday,  and 
having  to  do  its  Tuesday  over  again." 

This  line  is  now  little  used  by  sailors,  the  180th  meridian 
having  taken  its  place. 

The  consideration  of  this  line  in  the  solution  of  problems  in 
longitude  and  time  should  add  no  serious  difficulty.  Solve  the 
problem  completely,  leaving  out  of  account  the  date  line.  Then, 
if  the  time  of  the  given  place  is  west  of  the  line,  while  the  place 
whose  time  is  required  is  east,  we  simply  subtract  a  day,  and  if 
the  conditions  are  reversed,  we  add  a  day. 

I.  When  it  is  five  minutes  after  four  o'clock  on  Sunday 
morning  at  Honolulu,  what  is  the  hour  and  day  of  the  week  at 
Sydney,  Australia?  (Ray's  Higher  Arithmetic,  p.  iji,prob.  7.) 


THE  INTERNATIONAL  DATE  LINE. 


37 


II.  < 


1. 


x  HONOLULU 


157°  52  W.= 
longitude  of 
Honolulu. 


OREE«WU:M^*-T:- 
^3^ 


151°  11  E.= 
longitude  of 
Sydney. 

309°  3'=differ- 
ence  of  long- 
itude meas- 
ured from 
Honolulu 
through 
Greenwich 
to  Sidney. 

360°—  309°  3'  =  50°  57'=  difference  of  longitude 
measured  directly  on  the  equator  from  the  merid- 
ian through  Honolulu  to  the  meridian  through 
Sydney. 

15°=1  hr. 

1°=--  hr.=4  min. 


5. 
6. 

7.  50°  57"=50fJ°r=50i§0=50^X4  min.=3  hr.  23  min. 

48  sec.,  difference  of  time. 

8.  4  hr.  5  min.,  Sunday — 3  hr.  23  min.  48  sec.  =41  min. 

12  sec.,  Sunday. 

9.  Regarding  the  date  line,  Sunday  is  changed  to  Mon- 

day, since  Honolulu  is  east  of  the  line,  while  Syd- 
ney is  west  of  it. 

EXAMPLES. 

1.  When  it  is  5  o'clock  Monday  morning  at  Paris,  France,  longitude 
2°  20'  E.,  what  is  the  hour  and  day  of  the  week  at  Honolulu,  Hawaiian  Is- 
lands, longitude  157°  52'  W.? 

Ans.  19  min.  12  sec.  past  6  o'clock  P.  M.,  Sunday. 

2.  When  it  is  five  minutes  after  3  o'clock  on  Sunday  morning  at  Hon- 
olulu, Hawaiian  Islands,  longitude  157°  52'  W.,  whet  is  the  hour  and  day  of 
the  week  at  Sydney,  Australia,  longitude  151°  II7  E.? 

Ans.  41  min.  and  12  sec.  before  12  o'clock  P.  M.,  Sunday. 

3.  When  it  is  20  minutes  past  12  o'clock  on  Saturday  morning  at  Chi- 
cago, 111.,  longitude  87°  35',  what  is  the  hour  and  day  of  the  week  at  Pekin, 
China,  longitude  116°  26'  E.? 

Ans.  56  min.  4  sec.  past  1  o'clock  P.  M.,  Saturday. 

4.  When  it  is  ten  minutes  until  12  o'clock,  Friday,  midnight,  at  Con- 
stantinople, Turkey,  longitude  28°  59X  E.,  what  is  the  hour  and  day  of  the 
week  at  Honolulu,  Hawaiian  [slands,  longitude  157°  52X  W.? 

Ans.  22  min.  36  sec.  past  11  o'clock  A.  M.,  Friday. 

5.  At  what  hour  must  a  man  start,  and  how  fast  must  he  travel,  at  the 
equator,  so  that  it  would  be  noon  for  him  for  twenty-four  hours  ? 

Ans.  Noon  ;  1037.4  statute  miles  per  hr. 


38 


FINKEL'S   SOLUTION   BOOK. 


6.  What  is  the  difference  of  time  between  Constantinople,  Turkey, 
and  Sydney,  Australia?  Ans.  8  hr.  10  min.  48  sec. 

7.  A  traveler  sets  his  watch  with  the  time  of  the  sun  at  New  York. 
He  then  travels  from  there  and  on  arriving  at  his  destination  finds  that  his 
watch  is  1  hr.  20  min.  30  sec.  fast.     What  is  the  longitude  of  his  destination 
if  the  longitude  of  New  York  is  74°  (X  24"  W.?  Ans.  94°  7'  54"  W. 

8.  When  it  is  1  o'clock  P.  M.  at  Rome,  Italy,  longitude  12°  28'  E.,  what 
is  the  hour  at  New  York,  longitude  74°  Ox  24"  W.? 

Ans.  14  min.  G^-  sec.  past  7  A.  M. 

SOLUTIONS. 
Ex.  2. — Reduce  2  p.  3  pn.  1  tr.  1  hhd.  1  gal.  1  qt.  to  pints, 

126 
84 
42 
31* 

63    4      2 
bbl.     T.      hhd.     gal.   qt.   pt. 

1  11 

63         63 


P- 
2 
126 


pn. 
3 

84 


tr. 
1 
42 


252  gal.  252  gal.  42  gal. 


63  gal. 

42 
252 
252 


610  gal. 
4 


2440  qt. 


4882  pt. 

Ex.  3.     I.  Reduce  2  bu.  3  pk.  2  qt.  1  pt.  to  pints. 
Equation  Method. 

-1.     1  bu.=4  pk. 
2.     2  bu.=2x4  pk.=8pk. 
8  pk.+3  pk.=ll  pk. 

I  pk.=8  qt. 

II  pk.=llx8qt.=88  qt. 
88  qt.+2  qt.=90  qt. 

90  qt.=90x2  pt.=180  pt. 
180  pt.-f  1  pt.=181  pints. 
.-.  2  bu.  3  pk.  2  qt.  1  pt.=181  pints. 


Solution  :    II. 


Conclusion :     III. 


EXAMPLES. 


Solution:     II. 


Conclusion: 


Solution  :   II. 


Ex.  4.     I.   Reduce  529  pints  to  bushels. 
Equation  method. 

1.  2  pt.=l  qt. 

2.  529  pt.=529-4-2=264  qt.+l  pt. 

3.  8  qt.=l  pk. 

4.  264  qt.=264-7-8=33  pk. 

5.  4  pk.— 1  bu. 

6.  33  pk.=33-f-4=8  bu.+l  pk. 
III.  .-.  529  pints=8  bu.  1  pk.  1  pt. 

Ex.  5.     How    many    gallons   will  a  tank   4  ft.   long,  3   ft 
wide,  and  1  ft.  8  in.  deep  contain? 

'l.  4  ft.=length, 

2.  3  ft.=width,  and 

3.  1  ft.  8  in.=l£  ft.=depth. 

4.  4x3Xlf=20  cubic  ft.=contents  of  tank. 

5.  1  cu.   ft.=1728  cu.  in. 

6.  20  cu.  ft.=20Xl728  cu.  in.=34560  cu.  in. 

7.  231  cu.  in.=l   gal. 

8.  .;.  34560  cu.  in.=34560-f-231=149ff  gal. 

Conclusion  :  III.  .-.  The  tank  will  contain  149ff  gallons. 

(Fish's  Comp.  Arith.,  p.  126,  prob.  2.) 

EXAMPLES. 

1.  How  many  links  in  46  mi.  3  fur.  5  ch.  25  links? 

2.  How  many  acres  in  afield  containing  1377  square  chains? 

3.  How  many  cubic  inches  in  29  cords  of  wood? 

4.  In  1436  nails  how  many  Ell  English? 

5.  How  many  miles  in  3136320  inches? 

6.  In  47  ft).  2  1   3  3    1 3  19  gr.  how  many  grains  ? 

7.  Change   16   Ib.  3  oz.   1   gr.,  Troy    weight  to  Avoirdupois 
weight. 

8.  An  apothecary  bought  by  Avoirdupois  weight,  2  ft).  8  oz.  of 
quinine  at  $2.40   per  ounce,  which  he  retailed  at  20  ct.  a  scruple. 
What  was  his  gain  on  the  whole? 

9.  How  many  seconds  in  a  Dionysian  Period? 

10.  How  many  seconds  in  the  month  of  February,  1892. 

11.  How    many    seconds    in    the   circumference    of  a  wagon 
wheel? 

12.  How  long  would  it  take  a  body  to  move  from  the  earth  to 
the  moon,  moving  at  the  rate  of  30  miles  per  day. 

13.  If  a  man  travels  4  miles  per  hour,  how  far  can  he  travel  in 
2  weeks  and  3  days? 


40  FINKEL'S   SOLUTION   BOOK. 

14.  How  much  may  be  gained  by  buying  2  hogsheads  of  mo- 
lasses, at  40  ct.  per  gallon,  and  selling  it  at  12  cents  per  quart? 

Ans.  $10.08 

15.  In   74726807872   seconds,  how   many  solar   years? 

Ans.  2368  years. 

16.  At  $4    per  quintal,   how    many    pounds  of  fish   may  be 
bought  for  $50.24?  Ans.  1256  pounds. 

17.  How    many    bottles  of  3  pints  each  will  it  take  to  fill  a 
hogshead?  Ans.  168. 

18.  What   will  73  bushels  of  meal  cost,  at  2  cents  per  quart? 

Ans.  $46.72. 

19.  How  many  ounces  of  gold  are  equal  in  weight  to  6  ft),  of 
lead?  Ans.  87£oz. 

20.  What  is  the  difference   between  the  weight  of  42f  ft),  of 
iron  and  42.375  ft),  of  gold  ?  Ans.  52545  gr. 

21.  How  many  bushels  of  corn    will  a  vat    hold    that    holds 
5000  gallons  of  water.  Ans.  537  A  bu. 

22.  A  cellar  40  ft.  long,  20  ft.  wide  and  8  ft.  deep  is  half  full 
of  water.      What   will    it   cost   to   pump    it   out,  at  6  cents  a 
hogshead  ?  Ans.  $22.797+. 

23.  If  a  man  buys  10  bu.  of  chestnuts  at  $5  a  bushel,  dry  meas- 
ure, and  sells  the  same  at  25  cents  a  quart,  liquid  measure,  how 
much  does  he  gain?  Ans.  $43.09-f-  gain. 

24.  How  many  steps,  2  ft.  8  in.  each,  will  a  man  take  in  walk- 
ing a  distance  of  15  miles?  Ans.  29700. 

25.  How  many  hair's  width  in  a  40  ft.  pole,  if  48  hair's  width 
equals  1  line? 

26.  How  many  chests  of  tea,  weighing  24  pounds  each,  at  43 
cents  a  pound,  can  be  bought  for  $1548?  Ans.  150  chests. 

27.  How  long  will  it  take  to  count  6  million,  at  the  rate  of  80  a 
minute,  counting  10  hours  a  day?  Ans.  125  days. 

28.  How  long  will  it  take  to  count  a  billion,  at  the  rate  of  80 
a  minute,  counting  12  hours  a  day?  Ans.  -  - 

29.  What  will  15  hogsheads  of  beer  cost,  at  3  cents  a  pint. 

Ans.  $194.40. 

30.  How    many    shingles   will  it   take  to  cover  the  roof  of  a 
building  60  ft.  long  and  56  ft.  wide,  allowing  each  shingle  to  be  4 
inches  wide  and  18  inches  long,  and  to  lie   ^  to  the  weather? 

.  20160. 


31.  There  are  9  oz.  of  iron  in  the  blood  of  1  man.  How  many 
men  would  furnish  iron  enough  in  their  veins  to  make  a  plow- 
share weighing  22-J-  Ibs.  ?  Ans.  40- 


GREATEST   COMMON   DIVISOR.  41 

CHAPTER   VIII. 

GREATEST  COMMON  DIVISOR. 

1.  A  Divisor  of  a  number  is  a  number  that  will  exactly 
divide  it. 

2.  A    Common  Divisor  of  two   or  more   numbers  is  a 
number  that  will  exactly  divide  each  of  them. 

3.  The    Greatest    Common    Divisor 9  or    Highest 
Common  Factor,  of  two  or  more  numbers  is  the   greatest 
number  that  will  exactly  divide  each  of  them. 

I.     Find  the  G.  C.  D.  of  60,  120,  150, 180. 

II.     60=2X2X3X5. 
2.  120=2X2X2X3X5. 
3.  150=2X3X5X5. 
4.  180=2X2X3X3X5. 
5.  G.  C.  D.=2X3X5=30. 
III.  .-.  G.  C.  D.=30. 

Explanation. — By  inspecting  the  factors  of  each  number  we 
observe  that  2  is  found  in  each  set  of  factors;  hence,  each  of  the 
numbers  can  be  divided  by  2.  But  only  once,  since  it  is  found 
only  once  in  the  factors  of  150.  We  also  observe  that  3  will 
divide  the  numbers  only  once,  since  it  occurs  only  once  in  the 
factors  of  60  and  120.  Also,  5  will  divide  them  but  once,  since 
60,  120  and  180  contain  it  but  once.  Hence,  the  numbers,  60, 
120,  150, 180,  being,  divisible  by  2,  3  and  5,  are  divisible  by  their 
product,  2X3X5=30. 

I.     Find  the  G.  C.  D.  of  180,  1260,  1980. 

(1.     180=2X2X3X3X5. 
TT  J2.  1260=2X2X3X3X5X7. 
'  ]3.  1980=2X2X3X3X5X11- 

U.  G.  C.  D.=2X2X3X3X5=180. 
III.  .-.  G.  C.  D.  of  180,  1260,  1980=180. 

Explanation. — 2  being  found  twice  in  each  number,  they 
are  each  divisible  by  2x2  or  4  ;  also  3  being  found  twice  in  each 
number,  they  are  each  divisible  by  3x3  or  9.  5  being  found  in 
each  number,  they  are  each  divisible  by  5.  Hence,  they  are 
divisible  by  the  product  of  these  factors,  2X2X3X3X5=180. 

EXAMPLES. 

1.  Find  the  G.  C.  D.  of  78,  234,  and  468. 

2.  What  is  the  G.  C.  D.  of  36,  66,  198,  264,  600  and  720? 

3.  I  have  three  fields  :  the  first  containing  16  acres,  the  second 
20  acres,  and  the  third  24  acres.     What  is  the  largest  sized  lots 


42  FINKEL'S   SOLUTION   BOOK. 

containing  each  an  exact  number  of  acres,  into  which  the  whole 
can  be  divided?  Ans.  4  A.  lots. 

4.  A  farmer  has  12  bu.  of  oats,  18  bu.  of  rye,  24  bu.  of  corn 
and  30  bu.  of  wheat.     What  are  the  largest  bins  of  uniform  size, 
and  containing  an  exact  number  of  bushels,  into  which  the  whole 
can  be  put,  each  kind  by  itself,  and  all  the  bins  be  full? 

Ans.  6  bu.  bins. 

5.  A  has  a  four-sided  field  whose  sides  are  256,  292,  384,  and 
400  feet  respectively;  what  is  the  length  of  the  rails  used  to  fence 
it,  if  they  are  all  of  equal  length  and  the  longest  that  can  be 
used?  Ans.  4  ft. 

6.  In  a  triangular  field  whose  sides  are  288, 450,  and  390  feet 
respectively,  how   many  rails  will  it  require  to  fence  it,  if  the 
fence  is  5  rails  high,  and  what  must  be  the  length  of  the  rails  if 
they  lap  over  one  foot?  Ans.  Length  of  rail,  7  ft.  No.  940. 


CHAPTER   IX. 

LEAST    COMMON   MULTIPLE. 

1.     A.  Multiple  of  a  number  is  a  number  that  will  exactly 
contain  it ;   thus,  24  is  a  multiple  of  6. 

3.     A   Common,  Multiple  of  two  or  more  numbers  is  a 
number  that  will  exactly  contain  each  of  them. 

3.    The  Least  Common  Multiple  of  two  or  more  num- 
bers is  the  least  number  that  will  exactly  contain  each  of  them. 
I.     Find  the  L.  C.  M.  of  30,  40,  50. 

II.  30=2X3X5. 
2.  40—2X2X2X5. 
3.  50=2X5X5. 
4.  L.  C.  M.=2X2X2X3X5X5=600. 
III.  .-.  L.  C.  M.  of  30,  40,  50-=600. 

Explanation. — The  L.  C.  M.  must  contain  2  three  times,  or 
it  would  not  contain  40;  it  must  contain  5  twice,  or  it  would  not 
contain  50;  it  must  contain  3  once,  or  it  would  not  contain  30. 
Since  all  the  factors  of  the  numbers,  30,  40,  50,  are  contained  in 
th^  L.  C.  M.,  it  will  contain  each  of  them  without  a  remainder. 

I.     Find  the  L.  C.  M.  of  2310,  2,10,  30,  6. 

II.  2310=2X3X5X7X11- 
2.     210=2X3X5X7. 
3.       30=2X3X5. 
4.         6=2X3. 
5.  L.  C.  M.=2X3X5X7X  11=2310. 
III.  .-.  L.  C.  M.  of  2310,  210,  3(\  3 


LEAST  COMMON  MULTIPLE.  43 

Explanation. — 2  and  3  must  be  used,  else  the  L.  C.  M- 
would  not  contain  6.  2,  3,  and  5  must  be  used,  else  the  L.  C.  M- 
would  not  contain  30.  Hence  5  must  be  taken  with  the  factors 
of  6.  In  like  manner  7  must  be  taken  with  the  factors  already 
taken,  else  the  L.  C.  M.  would  not  contain  210.  The  factor  11 
must  be  taken  with  those  already  taken,  else  the  L.  C.  M.  would 
not  contain  2310.  Hence  2,  3,  5,  7,  and  11  are  the  factors  to  be 
taken  and  their  product  2310  is  the  L.  C.  M. 

I.  The  product  of  the  L.  C.  M.  of  three  numbers  between 
1  and  100  is  6804  ;  and  the  quotient  of  the  L.  C.  M.  divided  by 
the  G.  C.  D.  is  84.  What  are  the  numbers? 

1.  L.  C.  M.xG.  C.  D.=6804,  and 

L.C.M 

c  P  TT — 
3.  .-.'L.'C.'M.XG.  C.  D.-s-k'  C'  p=L.  C.  M.X 

G.  C.  D-X''=(G-  C.  D.  )2=6804-;-84=81. 


II. 


4.  G.  C.  D.— y'si  =9,  by  extracting  the  square  root. 


5.  .-.  L.  C.  M.=6804-i-9=756. 

6.  9=3x3. 

7.  756=2x2x3x3x3x7. 

8.  3x3x2x2=36. 

9.  3x3x3x2=54. 
10.  3x3x7      =63. 

III.     .-.  36,  54,  and  63=the  numbers. 

Explanation. — Since  9  is  the  G.  C.  D.,  each  of  the  numbers 
contains  the  factors  of  9.  Since  there  are  two  2's  in  the  L.  C. 
M.,  one  of  the  numbers  must  contain  these  factors.  In  like  man- 
ner one  of  the  numbers  must  contain  three  3's;  one  of  them  must 
also  contain  7.  .'•  We  write  two  3's  for  each  of  the  numbers,  two 
2's  to  any  set  of  these  3's,  and  3  and  7  with  either  of  the  remain- 
ing sets,  observing  that  the  product  of  the  factors  in  any  set  does 
not  exceed  100.  If  we  omit  2  in  step  9,  the  product  of  the  fac- 
tors is  27.  Hence  27,  36,  63  are  numbers  also  satisfying  the  con- 
ditions of  the  problem. 

EXAMPLES. 

1.  What  is  the  L.  C.  M.  of  13,  14,  28,  39,  and  42? 

2.  What  is  the  L.  C.  M.  of  6,  8,  10,  18,  20,  36,  and  48? 

3.  What  is  the  L.  C.  M.  of  18,  24,  36,  126,  20,  48,  96,  720, 
and  84? 

4.  What  is  the  smallest  sum  of   money  with  which  I  can 
purchase  a  number  of  oxen   at  $50  each,  cows  at  $40  each,  or 
horses  at  $75  each?  Ans.  $600. 


44 


FINKEL'S  SOLUTION   BOOK. 


5.  Find  three  numbers  whose  L.  C.  M.  is  840  and   G.  C.  D. 
42.  Ans.  84,  210,  and  420. 

6.  What  three  numbers  between  30  and   140  having  12  for 
their  G.  C.  D.  and  2772  for  their  L.  C.  M.  ?       Ans.  36,  84,  and  132. 

7.  At  noon  the  second,  minute,  and   hour  hands  of  a  clock 
are  together;   how  long  after  will  they  be  together  again   for  tho 
first  time? 

8.  J.  S.  H.  has  5  pieces  of  land;  the  first  containing  3  A. 
2  rd.  1  p.;  the  second,  5  A.  3  rd.  15  p.;  the  third  8  A.  29  p.v 
the  fourth,  12  A.  3  rd.  17  p.;  and  the  fifth,  15  A.  31   p.     Re- 
quired  the   largest  sized    house-lots,  containing    each   an  exact 
number  of  square  rods,  into  which  the  whole  may  be  divided. 

Ans.  1  A.  21  p. 

9.  The  product  of  the  L.  C.  M.  of  three  numbers  by  their 
G.  C.  D.==864,  and  the  L.  C.  M.  divided  by  the  G.  C.  D.=24; 
find  the  numbers.  Ans.  12,  18,  and  48. 


CHAPTER   X. 

FRACTIONS. 
1,     A    Fraction  is  a  number  of  the  equal   parts  of  a  unit. 

1.  Simple. 
1.  As  to  Form.<!  2.   Complex. 


2.     Fraction.  < 


'1. 

Common,  J 
or 
Vulgar. 

,2. 

As  to  Value." 

I 

i* 

Proper. 
Improper. 
Mixed. 

2. 

Decimal.  J 

'l. 
2. 
3. 

Pure. 
Mixed.            I 
Circulating.  <^ 

Pure. 
Mixed. 

3. 

Continued 

Fractions. 

3.  A  Common  Fraction,  or  Vulgar  Fraction,  is 

one  in  which  the  unit  is  divided  into  any  number  of  equal  parts; 
and  is  expressed  by  two  numbers,  one  written  above  the  other, 
with  a  horizontal  line  between  them.  Thus,  •§•  expresses  five- 
sixths. 

4.  A   Simple  Fraction  is  a  fraction  having  a   single 
integral  numerator  and  denominator;  as,  -f. 

5.  A    Complex  Fraction  is  a  fraction  whose  numer- 
ator, or  denominator,  or  both,  are  fractional;  as,  — ,   ^f,  -^. 

of     o^      5 


FRACTIONS.  45 

6.  A    CoMiptttt'fld   Fraction  is  a  fraction  of  a  fraction; 
as,  J  of  f  . 

7.  A   jProper    Fraction    is    a    simple    fraction    whose 
numerator  is  less  than  its  denominator;  as,  |. 

8.  An  Improper  Fraction  is  a  simple  fraction  whose 
numerator  is  greater  than  its  denominator;  as,  |. 

9.  A   Mixed   Wumber  is   a  whole   number  and   a    frac- 
tion; as,  3|. 

1C.  A  Decimal  Fraction  is  a  fraction  whose  denomi- 
nator is  ten,  or  some  power  of  ten;  as,  y3^,  T|¥,  T|^.  The  de- 
nominator of  a  decimal  is  usually  omitted  and  the  point  (.)  is 
used  to  determine  the  value  of  the  decimal  expression.  Thus, 


11.     A  Pure  Decimal  is  one  which  consists  of  decimal 
figures  only;  as,  .375. 

13.     A    Mixed    Decimal    is    one    which    consists    of   an 
integer  and  a  decimal;  as,  5.25. 

13.  A  Circulating  Decimal,  or  a  Circulate,  is  a  deci- 
mal in  which  one  or  more  figures  are  repeated  in  the  same  order; 
as,  .2121  etc.      When  a  common   fraction  is  in  its  lowest  terms 
and  the  denominator  contains  factors  other  than  2  or  powers  of 
2,  and  5  or  powers  of  5,  the  equivalent  decimal  fraction  will  be 

7 

circulating.     Thus,  T^Vir^o^  —  5  —  FT  wl^5    when    reduced    to    a 

L    X  o  X  o 

decimal,   be    circulating    because    the   denominator  contains  the 
factor  3. 

The  repeating  figure  or  set  of  figures  is  called  a  Repetend* 
and  is  indicated  by  placing  a  dot  over  the  first  and  the  last  fig- 
ure repeated. 

14.  A  Pure   Circulate  is  one  which  contains  no  figures 
but  those  which  are  repeated;  as,  .273. 


15.  A  Mixed   Circulate  is  one  which  contains   one  or 
more  figures  before  the  repeating  part;  as,  .45342. 

16.  A   Simple  Hepetend  contains  but  one  figure;    as,  .3. 

17.  A    Compound  Hepetend    contains    more    than  one 
figure;    as,  354. 

18.  Similar  Hepetends  are  those  which  begin  and  end 
at  the  same  decimal  places;  as,  .3467)  and  .0358- 

19.  Dissimilar  Repetends  are  those  which  begin  or 
end  at  different  decimal  places  ;  as,  .536.  .835,  and  .3567. 


46  FINKEL'S  SOLUTION  BOOK. 

2O.  A  Perfect  Hepetend  is  one  which  contains  as  many 
decimal  places,  less  1,  as  there  are  units  in  the  denominator  of  the 
equivalent  common  fraction  ;  thus,  7=.142857- 


21.  Conterminous  Jiepetends  are  those  which  end  at 
the  same  decimal  place;  as,   .4267,  -3275,  and   .0321. 

22.  Co-originous  Hepetends  are  those  which  begin  at 
the  same  decimal  place  ;  as,  .378,  -5624,  and  3-623. 

I.     Reduce  -      to  its  lowest  terms. 


Explanation.  —  Dividing  the  numerator  9,  by  3,  without 
changing  the  denominator,  the  value  of  the  fraction  is  dimin- 
ished as  many  times  as  there  are  units  in  the  divisor  3.  Dividing 
the  denominator  12,  by  3,  without  changing  the  numerator  9,  the 
value  of  the  fraction  is  increased  as  many  times  as  there  are  units 
in  the  divisor  3.  Hence,  if  we  divide  both  terms  by  3,  the  in- 
crease by  dividing  the  denominator  will  be  equal  to  the  decrease 
by  dividing  the  numerator,  and  the  value  of  the  fraction  will 
remain  unchanged. 

I.     Reduce  j-  to  a  higher  denomination. 


Explanation.  —  Multiplying  the  numerator  2,  by  4,  without 
changing  the  denominator,  the  value  of  the  fraction  is  increased 
as  many  times  as  there  are  units  in  the  multiplier  4.  Multiply- 
ing the  denominators,  by  4,  without  changing  the  numerator,  the 
value  of  the  fraction  is  decreased  as  many  times  as  there  are 
units  in  the  multiplier  4.  Hence,  if  we  multiply  both  terms  by 
4,  the  increase  by  multiplying  the  numerator  is  equal  to  the  de- 
crease by  multiplying  the  denominator,  and  the  value  of  the 
fraction  remains  unchanged. 

I.     Reduce  9£  to  an  improper  fraction. 

i.   9 


oi*  TT  J  2-     l=i=8-eighths. 

•  1  3.     9=9Xf  =-V-=9X8-eighths=r72-eighths. 

Conclusion-  III.     .'.  9%=^j>-=79-eighths. 

I.     Reduce  f  to  24ths. 

'  1.     f =f  f ,  or  8-eighths=24-twenty-fourths. 

2.     4=4  of  }J=A.  or   l-eighth=4  of  24- 
twenty-fourths=3-twenty-fourths. 

f  :=5  times  /j:^^! ,  or  5-eighths=5  times 
3-twenty-fourths=15-twenty-fourths. 
|=:||=r!5-twenty-fourths. 


II.  } 
3. 


FRACTIONS. 


47 


I.     Reduce  f  to  8ths. 

1.     f=|,  or  6-sixths=:8-eighths. 

9      i — i  ~f  8 f .a li,  or 

*•     t— £  01  f — o—  fi- 


ll. 


III. 
I. 


3.     4=5  times  ±?=- 


=J-  of  8- 

eighths. 

or  5-sixths=5  times 
IJ-eighths^Gf- 
eighths. 


Reduce     to  3rds. 


£Ji!  i4off^M*= 

I  O          O 


=    of 
3-thirds. 


III.    .-.  i~ 


Explanation. —  In  taking  \  of  f ,  we  must  divide  the  numer- 
ator by  2.     The  denominator  must  be  left  unchanged ;  for  that 
is  the  denomination  to  which  the  given  fraction  is  to  be  reduced. 
I.     Reduce  f  to  llths. 

1 .     f  ^rf^ll-elevenths. 

i i  _r  11 y~ ?j- i  Of  11-elevenths 

t-i  of  ii_n_n_^  _, 


II. 


-3  times 
S 


2i-elevenths. 

-^=3  times  24-elevenths 
~ll    =6felevenths. 


to  a  mixed  number. 


3-thirds=l. 

29-thirds=as  many  times  1  as  3- 

thirds  is  contained  in  29-thirds, 

which  is  9f  . 


III. 


I.     Reduce  f .  f ,  |  to  their  I,.  C.  Denominator. 

1.     L.  C.  D.=12. 

2. 

3. 

4. 

II.- 


48 


FINKEI/S  SOLUTION   BOOK. 
I.     Reduce   |,  |,  |  to  their  L.  C.  DenoMrntf-or 


fl.     L.  C.  D.=40. 

2.  !=«. 

li.  3.    4=4x1*=!*. 


i-  •'•  i,  I,  *=-**,  H 

I.     Add  |,  f,  J. 
1.  L.  C.  D.=24. 

2.  I=. 


4.  |=fxM= 
5-  J=x= 

6.  .'. 


I.     Reduce  f  to  a  fraction  whose  numerator  is  15. 


III.   .-.  f=if. 

I.     Reduce  f,  f ,  ^  to   equivalent   fractions   having 
least  common  numerators. 

1.  L.  C.  N.=12. 

2.  1=||. 

A 
4. 


in.  .-. 


I.      Subtract  f  from  - 
I.     L.  C.  D.=40. 

2-  1=4*- 

3-  f=fX«=it- 

4-  - 


TIL  ,.  A-t^U, 
1.     Multiply  J  by  f. 


fl- 

n.{2. 

13. 


f  X|=5  times 

III.     .'.       X*=£l: 


FRACTIONS. 


I.     Divide  -|  by  ^. 
II.  J  2.      |44=7  times  f=V- 

U.    4-HN4  of  ¥=44=H±. 


in.    ... 

Analysis  to  the  last  example  : 

'1.     nj-  is  contained  in  1,  or  J-,  7  times. 

2.  f  is  contained  in  l,orf,  ^  of  7  times— |-  times. 

3.  f  is  contained  in  l,  ^  of  J  times^-^T  times. 

A-     7  is  contained  in  -|,  5  times  ^  times,  or  |^J  times. 
Note. — By  inverting  the  divisor,  we  find  how  many  times  it  is 
contained  in  1. 

EXAMPLES. 

1.  One-fifth  equals  how  many  twelfths? 

2.  Reduce  -|,  ^,  -|,  and  \   to   fractions    having   a    common    de- 
nominator. 

1  ^ 

3.  Reduce  I-  to  a  fraction  whose  numerator  is  13.       Ans.  _  ! • 


4. 

5. 
6. 

7. 
8. 
9. 

10. 
11. 


i 
1 

1 
13.       l=what?     ^4  ns.  907200. 


Reduce  £  to  a  fraction   whose  denominator  is  11.     Ans.  _ 

11 

Reduce  ~|,  -$-,  f,  to  fractions  having  common   numerators. 
Add  £,  -J,  |l,  |,  and  ^. 
3  Of  8f— |  of  5— what? 
Multiply  f  by  8f . 
Multiply  I  of  9i  of  f  by  f  of  17. 
197      r;2 

1Z¥   .   Of 


Ans. 


a 

14.    !i= 


4 


50  FINKEL'S  SOLUTION   BOOK. 

15.     (2iX2*+*  of  TV)  X  (f)3-H7f-3iXff)=what? 


400000 
Ans'  407511: 


.  A+H-A       _ 

" 


-7H      62A  4f  -    '   4fx5i-200|^  "          3 

8 


17.  2-~-2-j-2-;-2H-2-r-2-r-2-r-2-:  i  :  |  :  ^  :  j 

Ans.  1. 

18.  Reduce  f  to  thirds.  Ans.  2f  thirds. 

19.  What  fraction  is  as  much  larger  than  f  as  f  is  less  than  f  ? 

Ans.  if. 

20.  What  is  the  value  in  the  13th  example  if  a  heavy  mark  be 
drawn  between  ^  and  -J-.  Ans.  If. 


21.     l--=what? 


22.  Subtract  *  of          from  ^  of  -j-  Ans.  f  JJJ. 

*«t  °f 

23.  What  is  the  relation  of  11  to  3? 

(1.       I=iof3. 

Solution:      4  2.     11=11  times  J  of  3=Jf  of  3=3§ 
(  times  3. 

Conclusion:     /.  11  is  3f  times  3. 

24.  What  is  the  relation  of  19  to  5?  Ans.  3J. 

25.  What  is  the  relation  of  T6T  to  24?  Ans.  ^ 

26.  What  part  of  3  is  2? 

Solution: 


i 
2.     2=2  times  J  of  3=§  of  3. 

Conclusion:     .-.  2  is  §  of  3. 

27.  What  part  of  6  is  7?  Ans.  J.. 

28.  What  part  of  3  is  J?  Ans.  TV 

29.  What  part  of  4  is  3?  Ans.  -^ 

30.  What  part  of  f  of  f  is  f  of  TV 


EXAMPLES.  51 

ri.    |oft=f 

Solution:        f     f  °f  Af-|V 

(  4.     ^  is  A  times  f  of  f =T72-  of  f . 
Conclusion:     .'.  f  of  T7^,  or  ^,  is  T7^  of  f  of  f . 

i  +  i,! 

31.  24i=what?  Ans.  1&. 

[Note. — This  is  a  continued  fraction.] 

32.  Find  the  number  of  which  75  is  ^. 

33.  Find  the  number  of  which  180  is  f . 

34.  if  is  f  of  what  number? 

11.     f  of  some  number=if . 
I  of  that  number=i  of  «=*. 
o.     I  oi  that  number,  or  the  number  rer 
quired,  =4  times  T%— if . 

Conclusion:     .-.  if  is  f  of  if. 

35.  27  is  .3  of  what  number?  Ans.  90. 

36.  f  of  if  is  \  of  ,f  of  what  number? 


*'  .    ''  '  f  =*  of  somj  nuniber  »  or 

4.  ^  of  some  number=§. 

5.  f  of  that  number,  or  that  number, 


Solution  : 


Conclusion:     .-.  §  of  |f  is  J-  of  f  of  -J¥5-. 

37.     A  watch  cost  $30,  and  this  is  f  of  f  of  the  cost  of  the 
watch  and  chain  together.     What  did  the  chain  cost.      Ans. 


38.  A  lost  f  of  his  money  and  then  found  f  as  much  as  he 
lost  and  then  had  $120;  how  much  money  had  he  at  first? 

39.  A  sum  of  money  diminished  by  f  of  itself  and  $6  equals 
$12;  what  is  the  sum?  Ans. 


40.  If  •£%  of  a  ton  of  hay  is  worth  $8^,  how  much  is  10  tons 
worth?  Ans.  $204. 

41.  What  number  is  that  if  of  which  exceeds  -fa  of  it  by 
111?  Ans.  216. 

42.  What  part  of  2J  feet  is  3J  inches?  Ans.  ^ 

43.  A  has  $2400;  f  of  his  money  plus  $500  is  }  of  B's;  what 
sum  has  B?  Ans.  $1600. 


52  FINKEL'S  SOLUTION  BOOK. 


44.     What  fraction  of      -^V— TT-^T  1     I  -^T+^^T  1 

Ans. 


45.  A  pole  stands  f  in  the  mud,  -^  in  the  water,  and  the  re- 
mainder, 12f  feet,  above  water.     Find  the  length  of  the  pole? 

Ans.  44f  feet. 

46.  If  48  is  ^  of  some  number,  what  is  f  of  the  same  num- 
ber? Ans.  63. 

47.  A  can  do  a  certain  piece  of  work  in  8  days,  and  B  can  do 
the  same  work  in  6  days.     In  what  time  can  both  together  do 
the  work?  Ans.  3^  days. 

48.  The  lesser  of  two  numbers  is  -     '  gQa  ,  and  their  differ- 

7  °*  °f 

ence  is  -  |-.     What  is  the  greater  number?  Ans.  -^-jp-- 

*r 

49.  What  number  multiplied  by  f  of  f  X3f  will  produce  |f  ? 

Ans.  f  . 

50.  What  number  divided  by  If  will  give  a  quotient  of  9J? 

Ans.  *$•• 

51.  A  post  stands  £  in  mud,  J  in  water,  and  21  feet  above 
the  water?     What  is  its  length?  Ans.  36  feet. 

52.  A  can  do  a  piece  of  work  in  8  days,  A  and  B  can  do  it  in 
5  days,  and  B  and  C  in  6  days.     In  what  time  can  A,  B,  and  C 
do  the  work?  Ans.  3T%  days. 

53.  If  f  of  6  bushels  of  wheat  cost  $4J,  how  much  will  f  of 
1  bushel  cost?  Ans.  80  cents. 

55.  What  number  diminished  by  the  difference  between  ^  and 
£  of  itself  leaves  1152?  Ans.  2268. 

56.  If  a  piece  of  gold  is  f  pure,  how  many  carats  fine  is  it? 

.     Ans.  15  carats. 

57.  The  density  of  the  earth  is  5f  times  that  of  water,  and 
the  sun  is  \  as  dense  as  the  earth.     How  many  times  denser  than 
water  is  the  sun?  Ans.       . 


CIRCULATING  DECIMAL: 


53 


CHAPTER    XI. 

CIRCULATING  DECIMALS. 
I.      Change   .63  to  a  common  fraction. 

.63=.636363+  etc.,  ad  injinitum. 

.636363+etc.,=.63+.0063+.000063+etc.,  ad  infinitum. 
13.     This  is  a  geometrical   infinite    decreasing  series    whose 
first  term  is  .63  and  ratio  .63-7-.0063=Tfo.     The  sum 

of  such  a  series  is =.63-r-(l — 5-^)=!-^=^-. 

III.   .-.  .63=T7T. 

I.     Reduce  1.001  to  a  common  fraction. 

1.  i.OOi=l.q0110011001100H0011+etc.,  ad  infinitum. 

2.  1.001=1.0011. 

3.  1.00li= l+.0011+00000011  +  000000000011+etc.,    ad 

infinitum. 


II., 


4.     .60li=rirst  term. 

5-     TTmnr=-001  l-r-00000011=ratio. 

6.     . 


17.      .'. 

(Ray's  H.  A.,  p.  120,  ex.  8.) 

III.   /.  1.001=1*^. 

Remark. — Since  the  denominator  of  the  ratio  is  always  ten  or 
some  power  of  ten,  the  numerator  of  the* fraction  resulting  from 
subtracting  the  ratio  from  1,  will  have  as  many  9's  in  it  as  there 
are  ciphers  in  the  denominator  of  the  ratio.  By  dividing  the  first 
term  by  this  fraction,  its  numerator  becomes  the  denominator  of 
the  fraction  required.  Hence,  a  circulate  may  be  reduced  to  a 
common  fraction  by  writing  for  the  denominator  of  the  repetend 
as  many  9's  as  there  are  figures  in  the  repetend.  Thus,  .63— 


I.     Reduce  .034639  to  a  common  fraction. 
i        034639  -  034«  *  »-34ttl.-34X  999  +639_  34  X  999+639 
~-TO4ttf=3!Tooo  VWr  1000X999   ' 

__  34  X  (1000—1)  +639       34000—  34+639       34000+639—  34 

9990<X)  999000  999000 

34639—  34_  34605  __  6921  =  2807         769 
999000    ~999000~199800      66600  ~22200' 


54  FINKEL'S  SOLUTION  BCQK. 

In  case  the  circulate  is  mixed,  we  have  the  following  rule  : 

1.  For  the  numerator,  subtract  that  part  which  precedes  the 
repetend  from  the  whole  expression,  both  quantities  being  consid- 
ered as  units. 

2.  For  the  denominator,  write  as  many  9*s  as  there  are  figures 
in  the  repetend,  and  annex  as  many  ciphers  as  there   are  decimal 
fgures  before  each  repetend. 

I.     ADDITION   OF   CIRCULATES. 

I.     Add  5.0770,  .24,  and  7.124943. 

(1.     5.0770    =  5.0770      =  5.07707707  etc. 
2.       .24        =    .242        =     .24242424  etc. 
3.     7.124943=  7.124943J=  7.124943J2  etc. 
III.  .-.  Sum=12.44  12.4444444  etc.=12.44. 

Explanation. — The  first  thing,  in  the  addition  and  subtrac- 
tion of  circulates,  is  to  make  the  circulates  co-originous ,  /.  e. ,  to 
make  them  begin  at  the  same  decimal  place.  That  is,  if  one  be- 
gins at  (say)  hundredths,  make  them  all  begin  at  hundredths, 
providing  that  each  circulate  has  hundredths  repeated.  It  is 
best  to  make  them  all  begin  with  the  circulate  whose  first  re- 
peated figure  is  farthest  from  the  decimal  point,  though  any 
order  after  that  may  be  taken.  In  the  above  example  we  have 
made  them  all  begin  at  hundredths.  After  having  made  them 
all  begin  at  hundredths,  the  next  step  is  to  make  them  con- 
terminous, i.e.)  to  make  them  all  end  at  the  same  place.  To 
do  this,  we  find  the  L.  C.  M.  of  the  numbers  of  figures  repeated 
in  each  circulate,  then  divide  the  L.  C.  M.  by  the  number  of  fig- 
ures repeated  in  each  circulate  for  the  number  of  times  the  figures 
as  a  group  must  be  repeated.  Thus,  the  number  of  figures  in  the 
first  repetend  is  3;  in  the  second,  2;  and  in  the  third,  6. 

The  L.  C.  M.  of  3,  2,  and  6  is  6.       6-r-3=2.      .*.  770  must 
be    repeated    twice.       6-j-2=3.     .'.    42    must  be   repeated    three 
times.     6-£-6=l.     .'.   249431  must  be  taken  once. 
I.     Add  .946,  .248,  5.0770,  3.4884,  and  7.124943. 

fl.       .946       =     .946  =     .946666666666666  etc. 

2.  .248       =     .2484  =     .2484-84848484848  etc. 

3.  5.0770    =  5.07707        =  5.077077077077077  etc. 
IL<U.     3.4884     =  3.488448       ==  3.488448844884488  etc. 

5.  7.124943=   7.12494312  =  7.124943124943124  etc, 

6.  Sum  =16.88562056205620+,    • 
=16.885620. 

III.     .-.  Sum=16.885620. 


III. 
I. 

II. 
III. 

I. 


I. 


CIRCULATING  DECIMALS. 


II.     SUBTRACTION  OF  CIRCULATES. 
Subtract  190.476  from  199.6428571 

1.  199.6428571  =  199.64285714 

2.  190.476          =  19Q  47619047 


55 


I     Difference     =      9.1666666  =  9.16. 
.-.     Difference=9.16. 
Subtract  13.637    from    104.1. 
1.     104.1     =104.14  =104.1414141  etc. 
13.637=  13.637=  13.6376376  etc. 


-3. 


Difference  =  90.503776 

.-.     Difference=90.503776. 

III.     MULTIPLICATION  OF  CIRCULATES. 

Multiply  .07067  by  .9432. 
.07067=.070677 
.9432  =       .9^?-         Multiply  by  the  fraction  thus : 

.06§609  .070677 

.003056  16 

.066665=product.       .42406=.424062 
.7067  =.706770 

37)1.13083(3056= 

HI  003056,  be- 
cause the 
fraction  is 


208 
185 


Multiply  1.256784  by  6.42081. 

1.256784=1.2567842 

6.42081  =   6.420T9T 

.02513568  =  .0251356851 

.5027137  =  .502713702 7 

7.540705  —  7.5407055407 

.001028270  =  .001028276° 


233 
222 

~ 


Multiply  by  the 
fraction  thus  : 

1.2567842 


8.069583198 


11).011311Q57 
.001628276- 


56  FINKEL'S  SOLUTION  BOOK. 

Remark. — In  multiplying  by  any  number,  begin  sufficiently  far 
beyond  the  last  figure  of  the  repetend,  so  that  if  there  is  any  to 
carry  it  may  be  added  to  the  repetends  of  the  partial  products, 
making  them  complete.  Thus  in  the  above  example,  when  mul- 
tiplying by  4,  we  begin  at  5,  the  second  decimal  place  beyond  4,  the 
last  figure  of  the  repetend  ;  and  so  when  we  multiply  4  by  4,  thtf 
first  figure  of  the  repetend  in  the  partial  product  is  7. 

IV.     DIVISION  OF  CIRCULATES. 

RULE. —  Change  the  terms  to  common  fractions;  then  divide  as 
4n  division  of  fractions,  and  reduce  the  quotient  to  a  repetend. 

I.  Divide  .75  by  .1 

fl.     .75=H=» 

II.  2.     .1=4. 

[3.     f-f  -Hr=ifx9=fi=6.8181  etc.=6.81. 

III.  .-.  .75-r-.i=6.8i. 

EXAMPLES. 

1.  Add  .87,  -8,  and  876.  Ans.  2.644553. 

2.  Add  .3,  .45,  .45,  .351,  .6468,  .6468,  .6468,  and  6468. 

Ans.  4.1766345618. 

3.  Add  27.56,  5.632,  6.7,  16.356,  .71,  and  6.1234. 

Ans.  63.1690670868888. 

4.  Add  5.16345,  8.6381,  and  3.75. 

Ans.  17.55919120847374090302. 

5.  From  315.87  take  78.0378.  Ans.  237-838072095497. 
•6.     From  16.1347  take  11.0884.  Ans.  5.0462. 

7.  18  is  .6  of  what  number?  Ans.  27. 

8.  From  ^  take  ^.  Ans.  .1764705882352941. 

9.  From  5.12345  take  2.3523456. 

-4*5.2.7711055821666927777988888599994. 

10.  Multiply  87.32586  by  4.37.  Ans.  381.6140338. 

11.  Multiply  382.347  by  .03.  Ans.  13.5169533. 

12.  Multiply  .9625668449197860  by  .75.  Ans.  .72. 

13.  Divide  234.6  by  .7.  Ans.  701.714285. 

14.  Divide  13.5169533  by  3.145.  Ans.  4.297. 


PERCENTAGE  AND  ITS  VARIOUS  APPLICATIONS.  57 

15.  Divide  2.370  by  4.923076.  Ans.  481. 

16.  Divide  ,36  by  .25.  Ans.  1.4229249011857707509881. 
17-  Divide,  .72  by  .75.                            Ans.  .9625668449197860. 

18.  54.0678132-r-8.594=what?  .  Ans.  6.290. 

19.  4.956-f-.75=what?  Ans.  6.6087542. 

20.  7.714285-7-.952386=what?  Ans.  8.1. 


CHAPTER  XII. 

I.     PERCENTAGE  AND  ITS  VARIOUS  APPLICATIONS. 

1.  Percentage  is  a  method  of  computation  in  which  100  is 
taken  as  the  basis  of  comparison. 

2.  Per  cent,  is  an  abbreviation  from  the  L,atin,  per  centum, 
per,  by,  and  centum,  a  hundred. 

3.  The  Terms  used  in  percentage  are  the  Base,  the  Rate, 
the  Percentage,  and  the  Amount  or  Difference. 

4.  The  Base  is  the  number  on  which  the  percentage  is 
computed. 

5.  The  Rate  is  the,  number  of  hundredths  of  the  base 
which  is  to  be  taken. 

6.  The  Percentage  is  the  result  obtained  by  taking  a  cer- 
tain per  cent,  of  the  base. 

7.  The  Amount  or  Difference  is  the  sum  or  difference 
of  the  base  and  percentage. 

8.  The  sign,  %,  is  used  instead  of  the  words  i(per  cent." 
and  "one-hundredths,"  following   the   number   expressing   the 
rate.     Thus,  for  example,  for  5  per  cent.,  or  5  one-hundredths, 
we  write  5%. 

Hence,  we  have  the  following  identical  expressions: 

5  per  cent. =5  one-hundredths=yf^=.05^5%.  In  each  of 
these  expressions  the  fractional  unit  is  TJ^.  The  fundamental 
principle  of  percentage  is  that  our  computation  shall  be  made  on 
the  basis  of  hundredths.  That  this  principle  be  not  violated, 
the  denominator  of  the  fraction  must  always  be  100.  Thus, 
since  TII)%=TIQ,  we  can  take  T^  of  a  number  instead  of  ^  °f  it 
and  get  the  same  result;  but  using  fractions  whose  denomina- 
tors are  numbers  other  than  100  to  express  the  rate  is  not  the 
method  of  percentage,  but  merely  the  method  of  common  frac- 
tions. However,  in  teaching  percentage  the  method  of  common 


58  FINKEL'S  SOLUTION  BOOK. 

fractions  should  also  be  used,  as  this  method,  because  of  its 
brevity,  is  more  often  used  in  practice. 

As  an  illustration,  find  5%  of  $600. 

1.  100  one-hundredths,  or  {%%,  or  1.00,  or  100%=$600, 

2.  1  one-hundredth,  or  -j-fr0  ,  or  .01,  or  1%,=^  of 
II.  ^j  $600=$6, 

3.  5  one-hundredths,  or  T^,  or  .05,  or  5%—  5  times 


III.     /.  5%  of  $600=$30. 
I.     What  is  8%  of  150  yards? 


FIRST   SOLUTION. 


I.    m=lW  yards. 
II.     2.     T^Tfo  of  i$r=Tiir  of  150  yards=1.5  yards. 

3.     y^¥=:8  times  1.5  yards=12  yards. 
III.     /.8%  of  150  yards=:12  yards.   " 

Remark.  —  This  solution  is  by  the  method  of  percentage  purely. 


II.  \  2. 
(3. 


SECOND   SOLUTION. 


1.     100%=:150  yards. 

1%=-^  of  100%=^  of  150  yards=1.5  yards. 
8%=8  times  1%—  8  times  1.5  yards=12  yards. 


III.     /.8%  of  150  yards=12  yards. 

THIRD   SOLUTION. 

Briefly,  by  fractions  : 

8%  of  150  yards^y^  of  15(0  yards=12  yards. 


CASE   I. 

P—  tage. 

Formula.—  BXR=P,  where  B  is  the  base,  R  the  rate,  and 
P  the  percentage. 


PERCENTAGE  AND  ITS  VARIOUS  APPLICATIONS.  59 

I.      What  is  8%  of  $500? 

rl.     100%— $500, 
II J  2.         1  %— y^  of  $500— $5,  and 

U.          8%—  8  times  $5— $40. 
III.     .-.     8%  of  $500— $40. 

1.      What  is  |%  of  800  men? 
-1.      100%— 800  men. 

1%—  YTFTT  °f  800  men— 8  men,  and 
^%=|  times  8  men— 6  men. 
III.     .-.     |%  of  800  men— 6  men. 


II./2. 
U. 


II 


I.     What  is  10%  of  20%  of  $13.50? 

1.50. 

2.        l%=Ti<>  of  $13.50— $.135,  and 
.3.     20%— 20  times  $.135— $2.70. 


(2.)        100%— $2.70. 


(3.)  l%=riir  of  $2.70— $.027,  and 

(4.)  10%— 10  times  $.027— $.27— 27  cents. 

III.  .-.  10%  of  20%  of  $13.50— 27  cents. 

I.  A.  had  $1200  ;  he  gave  30%  to  a  son,  20%  of  the  remain- 
der  to  his  daughter,  and  so  divided  the  rest  among  four 
brothers  that  each  after  the  first  had  $12  less  than  the 
preceding.  How  much  did  the  last  receive? 

rl.    100%— $1200, 
J2.        1%—  Tfo  of  $1200— $12,  and 
(     ')  3.     30%— 30  times  $12— $360— son's  share. 
U.  $1200— $360— $840— remainder, 
rl.  100%— $840. 

I  2.        1%=1^  of  $840— $8.40,  and 

(2.  W3.     20%— 20  times  $8.40— $168— daughter's  share. 
j4.  $840— $168— $672— amount   divided   among   four 

brothers. 

(3.)         100%— fourth  brother's  share; 
(4.)         100%+$12=--third  brother's  share. 
(5.)         100% +$24— second  brother's  share,  and 
(6.)         100%  +  $36=first  hi  other's  share. 
(7.)        100%+(100%+$12)-j-(100%+$24)+(100%+ 
$36)— 400%+$72— am'tthe  four  brothers  rec'd. 
(8.)         $672— amount  the  four  brothers  received. 
(90        ..400%-f$72— $672. 
( 10. )            400  %  —$672— $72— $600. 
/ 1 1  \  ~\  ot        i     f\f  'tfinri *ki  ^n 

IXJL.J  -I   m TlTlf  <pOUU •pl.c'v/. 

(12.)         100%— 100   times   $1.50—  $150—  fourth    brother's 
share. 

III.  .-.  The  last  received  $150.  (R.  H.  A.,  p.  191,prob.  25.) 


60  FINKEL'S  SOLUTION  BOOK. 

1.     What    number    increased    by  20%  of  3.5,  diminished  by 
12^%  of  9-6,  gives  3|? 

100%=the  number. 
.   100%  =3.5, 

;.        1%=T^-  of  3.5=.035,  and 
.3.     20%  =20  times  .0*5=.7. 
TTJ         rl.  100%  =9.6, 

(3.)<|2.       1%=TF(T  of  96=.096,  and 

12£%=12-J  times  .096=1.2. 
(4.)     .-.  100%+-7— 1.2=3i, 
(5.)         100%— .5=3.5,  and 
1(6.)         100%=4,  the  number. 
III.   .-.  The  number=4.  (/?.  H.  A.,  p.  191,  prob.  26.} 

CASE  II. 

Given  j  the  base  *nd  the  I  to  find  the  rate  per  cent. 
(       percentage       j 


Formula. — P^B=R,  where  B  is  the  base,  P  the  percentage 
and  R  the  rate  per  cent. 

I.      750  men  is  what  %  of  12000  men? 

12000  men=100%, 

2.  1  man=T2^o  of  100%=Tf  ff  %  ,  and 

,3.          750  men=750  times  -&*%*=&\%- 
III.      .'.   750  men  is  6^%  of  12000  men. 

I.      A's  money  is  50%  more  than  B's;  then  B'sishow  many  % 
less  than  A's? 

II.    100%=B.'s  money.     Then, 
2.   100%+50%=150%=A.'s  money. 
3.   150%=100%  of  itself. 
4.        1  %=Tio  of  100%=|%  ,  and 
5.     50%=50  times  f  %=33£%. 
III.      .-.  B.'s  money  is  33^%  less  than   A.'s 

(R.  H.  A.,p.  192,  prob.  11.) 

I.      30%  of  the  whole  of  an  article  is  how  many  %  of  f  of  it? 

'1.    100%=whole  article. 
2.  66f  %=f  of  100%=f  of  the  article. 
?=100%  of  itself. 


4.  1%=—  of  100%=H%,  and 

bo  3 

5.  30%=30  times  U%=45%. 


Ill       .-.  30%-of  the  whole  of  an  article  is  45%  of  f  of  it, 

(R.  H.  A.,  p.  192,  prob.  20.) 


I. 


PERCENTAGE. 


61 


If  a  miller    takes  4  quarts  for  toll   from  every  bushel  he 
grinds,  what  %  does  he  take  for  toll? 

(1.   1  bu.—  32  qt. 
TT  1  2.  32  qt.—  100%, 
'13.   1     t.= 


V4.  4  qt.=4  times  3^%—  12|%. 
III.     .'.  He  takes  12|%  for  toll. 


CASE  III. 


~  .         (  tne  percentage  and  )  .    r    ,  ,  ,     , 
Glven  to  find  the  base' 


I  the  rate  per  cent 


d  )  . 
.  }  to 


Formula.—  P-^R—B,  where  P  is  the  percentage,  R  the  rate 
per  cent.,  and  B  the  base. 


$24  is  f%  of  what  sum? 

1.  100%=sum. 

2.  |%=$24, 

3.  i%=4  of  $24=48, 

4.  ,  or  1%,=8  times  $8=$64,  and 


III. 


I. 


5.     100%—  100  times  $64—  $6400. 
.-.  $24isf%  of  $6400. 

I  drew  48%  of  my  funds  in  bank,  to  pay  a  note  of   $150; 
how  much  had  I  left? 


III. 


100%— amount  in  bank. 

48%— amount  drawn  out. 
100%— 48%— 52%— amount  left. 

48%— $150, 
!%=&  of  $150— $3.125,  and 

52%— 52  times  $3.125— $162.50— amount  left. 

$162.50— amount  I  had  left. 


III. 


I  pay  $13  a  month  for  board,  which  is  20%  of  my  salary; 
what  is  my  salary? 

100%—  my  monthly  salary. 

20%—  $13, 

l%=^r  of  $13—  $.65,  and 
100%—  100  times  $.65—  $65,  my  monthly  salary. 

.'.   $780—12  times  $65—  my  yearly  salary. 

My  salary=$780.  (R.  H.  A,,  p.  194,  prob.  20.) 


I. 
2. 
3. 
4. 
5. 


62 


FINKEL'S  SOLUTION  BOOK. 


CASE   IV. 


Given 


\  the       °Unt 

( 


to  find  the  base. 


, 
rate  per  cent. 

Formula.  —  A-*-(\+R)=B,  where  A  is  the  amount,  that  is, 
the  base  and  the  percentage,  R  the  rate  per  cent.,  and  B  the 
base. 

III.     .-.  $540  is  8%  greater  than  $500. 

I.     A  sold  a  horse  for  $150  and  gained  25%;    what   did   the 
horse  cost? 

1.  100%=cost  of  horse. 

2.  25%=gain. 

3.  100%+25%=125%=selling  price  of  horse,  and 

4.  $150=selling  price  of  horse  ; 

.'.  125%=$150, 

of  $150=$1.20,  and 


III. 


5. 
6. 
7.  100%=100  times  $1.20=412Q=cost  of  horse. 

.-.  The  horse  cost  $120. 


I.  I  sold  two  horses  for  the  same  price,  $150;  on  one  I 
gained  25%  and  on  the  other  I  lost  25%;  what  was  the 
cost  of  each? 

1.  100%=cost  of  first  horse. 

2.  25%=gain. 

3.  100%+25%=125%=selling  price  of  first  horse, 
A.  <4.     $150=selling  price  of  first  horse; 

5.  .'.  125%=$150, 

6.  1%=T^.  of  $150=$1.20,  and 

7.  100%=100  times  $1.20=$120=cost  of  first  horse. 
'1.     100%=cost  of  second  horse. 

2.  25%=loss  on  second  horse. 

3.  100% — 25%=75%=selling  price  of  2d  horse,  and 
B    <J4.     $150=selling  price  of  second  horse; 

5.  .-.  75%=$150, 

6.  1  %=TV  of  $150=$2,  and 

7.  100%=100  times  $2— $200=cost  of  second  horse. 


TTT 
11 


J$120=cost  of  first  horse,  and 
'*  1  $200=cost  of  second  horse. 


I.     A  coat  cost  $32;  the  trimmings  cost  70%   less,  and  the 
making  50%  less  than  the  cloth;  what  did  each  cost? 


PERCENTAGE. 


63 


II. 


III. 


II. 


III. 


r  1. 

2. 
3. 
4. 
5. 
6. 

8. 

9. 

10. 


100%=cost  of  cloth.     Then 
100%—  70%=30%=cost  of  trimmings,  and 
100%—  50%=50%=cost  of  making. 
100%+30%+50%=180%=cost  of  coat. 
$32=cost  of  coat; 


^-       of  $32=$. 


" 


IOO%=tlOO  times  $.1777£=$17.77£=cost  of  cloth. 
30%=30  times  $.1777J=$5.33i=cost  of  trimming. 
50%=50  times  $.1777J=$8.88f=cost  of  making. 

r$17.77J=cost  of  cloth, 

<[$  5.33^=cost  of  trimmings,  and 

^$  8.88|=cost  of  making. 

(R.  H.  A., p.  196,prob.  12.) 


n  a  company  of  87,  the  children  are  37^%  of  the  women, 
who  are  444%  of  the  men;  how  many  of  each? 


(1.) 

(2.) 


(4-) 

(5.) 
(6.) 

(V.) 

(8.) 

(9.) 

.(10.) 


100%— number  of  men.     Then 
44f%=number  of  women. 


3.  37^%=37|-  times  .44f  %=16f  %=number  of  chil- 
dren in  terms  of  the  number  of  men. 

10C%+44f%-f-16f%  =  161^-%=  number  in  the 
company, 

87— number  in  the  company  ; 


l%=i        of  87=.54, 

100%-=100  times  .54==54=number  of  men, 
444%==44£  times  .54=24=number  of  women,  and 
16|%=16|  times  .54=19=number  of  children. 


54=number  of  men, 
24=number  of  women,  and 
19=number  of  children. 


H.  A.,  p.  197,prob.  20.) 


I.  Our  stock  decreased  33^%,  and  again  20%;  then  it  rose 
20%,  and  again  33^-%;  we  have  thus  lost  $66;  what 
was  the  stock  at  first? 


64  FINKEL'S   SOLUTION  BOOK. 


100%=original  stock. 

33i%=decrease. 

100%—  33£%=66f  %=stock  after  first  decrease. 

100%=66f%, 

!%=TiTof  66J%=f%,  and 
20%—  20  times  f  %=13£%=second  decrease. 
66f  %—  13i%=53£%=stock  after  second  decrease. 


20%=20  times  .53£%=10f  %—first  increase. 
II.<|  U.  53^%+lOf  %=64%=stock  after  first  increase 

2.       1%=T^  of  64%=.64%,  and 
I  33£%=33|  times  .64%=21£%=second  increase. 
[.     64%-f-21£%=85i|-%=stock  after  second  increase. 

100%—  85^%=14f  %=whole  loss; 

$66=whole  loss; 

...  14|  %= $66; 

HO.)         I%=:rr5  of  $66— $4.50,  and 
\       /  /       14 

(11.)         100%=100  times  $4.50=$450=original  stock. 
III.  .-.  $450=original  stock. 

I.  A  brewery  is  worth  4%  less  than  a  tannery,  and  the  tan- 
nery 16%  more  than  the  boat;  the  owner  of  the  boat 
has  traded  it  for  75%  of  the  brewery,  losing  thus  $103 ; 
w'nat  is  the  tannery  worth  ? 

FIRST    SOLUTION. 

(1.)        100 %=value  of  the  tannery.     Then 
(2.)        100% — 4%=96%=value  of  the  brewery. 

'1.  100%=value  of  the  boat.     Then  [the  boat. 

2.  I00%+16%=116%=value  of  tannery  in  terms  of 

3.  116%=100%,  the  value  of  tannery  from  step  (1), 

4.  l%=nhr  of  100%=ff  % ,  and 

5.  100%=100  times  jfr%=86ff%=v&lue  of  the  boat 

in  terms  of  the  tannery. . 
\.  100%=96%, 
llA    (A  \J2-       1%=  T*irof96%=.96%,  and 

75%=75  times  .96%=72%=what    the    owner 

of  the  boat  received  for  it. 

•'•  862%%— 72%— 14^V%=what  the  owner  of  the 
boat  lost  in  the  trade. 
(6.)         $103=what  he  lost; 
(7.)         ...  14^g-%=:$103, 

(8.)         l^^L-of  $103— $7.25,  and 

(9.)         100%=100  times  $7.25=$725=value  of  tannery. 
III.     .'.  $725=value  of  the  tannery.     (R.H.A.,p.W7,prob.2S.) 


(3.) 


PERCENTAGE. 


65 


Remark. — The  value  of  the  brewery  and  boat  being  ex- 
pressed vi  terms  of  the  tannery,  75%  of  the  brewery  is  also  ex- 
pressed in  terms  of  the  tannery;  hence,  it  is  plain  that  the  owner 
of  the  boat  has  traded  86^69%  for  72%  of  the  same  value,  losing 

86/^%— 72%,  or  14^%. 

SECOND    SOLUTION. 

(1.)         100%— value  of  the  boat.     Then 

(2.)         100%+16%— 116%—  value  of  the  tannery. 

1(1.  100%— 116%, 
(3.K2.       1%=^  of  116%— 1.16%,  and 
l3.       4%—  4  times  1.16%— 4.64%. 
(4.)         116%— 4.64%— 111.36%— the  value  of  brewery  in 
terms  of  the  boat, 
rl.  100%— 111.36%, 
\\\    ,,.  J2.       1%—^  of  111.36%— 1.1136%,  and 

I')  3.     75%— 75    times     1.1136%— 83.52%— what    the 
owner  of  the  boat  received  for  it. 
(6.)         .-.  100%— 83.52%— 16.48%— what  he  lost  in  the 
trade. 
(7.)         $103— what  he  lost. 
(8.)        .',  16.48%— $103, 
(9.)         l%=nr?frir  of  $103— $6.25,  and 
.(10.)         116%— 116  times  $6.25— $725— value  of  tannery. 
III.     .-.  $725— value  of  the  tannery. 

THIRD    SOLUTION. 

100%— value  of  brewery. 

100%— value  of  tannery.     Then 

100%—  4%—  96%— value  .of  the  tannery. 

.-.  96%.— 100%,  the  value  of  brewery  in  step  (1), 

l%=-fa  of  100%— 1.041%,  and 

100%— 100  times  1.04^%— 104£%— value  the  tan- 
nery in  terms  of  the  brewery. 

100%— value  of  boat.     Then 

100%+16%— 116%—  value  of  the  tannery  in 
terms  of  the  boat. 

.-.  116%— 104^%,  the  value  of  the  tannery  in  step 
5  of  (2), 
1%— rle  of  104^%— .89^1%  ,  and 

100%— 100  times  .89if£%— 89  jf£%— value  of  the 
boat  in  terms  of  the  tannery,  and  consequently 
in  terms  of  the  brewery. 

.*«  89-H1  %—  75%— 14iff  %—  what  the  owner  of 
the  boat  lost  in  the  trade. 

$103— what  the  owner  of  the  boat  lost ; 


(2.) 


(3.) 


(4.) 

(5.) 
(6.) 

(7.)         i%=_L^  of  $103=16.96,  and 

I        /\  •  M 

I  (8.)       104i%— 104^  times  $6.96— $725=value  of  tannery. 


66  FINKEL'S  SOLUTION  BOOK. 

III.  .-.  $725=value  of  of  the  tannery. 

Remark. — In  step  5  of  (3),  we  have  the  value  of  the 'boat  in 
terms  of  the  tannery  ;  but  the  value  of  the  tannery  is  in  terms  of 
the  brewery:  hence,  the  value  of  the  boat  is  also  in  terms  of  the 
brewery.  The  owner  of  the  boat,  therefore,  traded  89|ff  %  for 
75%  of  the  same  value. 

MISCELLANEOUS    PROBLEMS. 

I.     A  man  sold  a  horse  for  $175,  which  was  12-^-%   less  than 
the  horse  cost;  what  did  the  horse  cost? 

1.  100%— cost  of  horse. 

2.  12|%— loss. 

3.  100%— 12|%— 87i%— selling  price. 
,  4.  $175— selling  price. 

IL<>5.  .-.  87i%— $175, 

6.  1^=_L  Of  $175— $2,  and 

7.  100%— 100  times  $2— $200, 

III.     .-.  $200— cost  of  the  horse.  (R.  3d  p.,  p.  204,  Pr°t>-  5.) 

I.     A  miller  takes  for  toll  6  quarts  from   every  5  bushels  of 
wheat  ground;  what  %  does  he  take  for  toll.? 

1.  1  bu.— 32  qt. 

2.  5bu.— 5  times  32  qt.— 160  qt. 


II. 


4.  160  qt.— 100% 


c, 


4.  1  qt.— T^  of  100%— f%,  and 

5.  6qt.—6  times |%—3f%. 

III.  "  .-.  He  takes  3f  %  for  toll.  (R.  3d  p.,  p.  204,  prob.  H-) 

I.     A  farmer  owning  45%  of  a  tract  of  land,  sold   540   acres, 
which  was  60%  of  what  he  owned;  how  many  acres  were 
there  in  the  tract? 
(1.)         100%— number  of  acres  in  the  tract. 

1.  100%— numbers  of  acres  the  farmer  owned. 

2.  60%=number  of  acres  the  farmer  sold. 

3.  540  acres—what  he  sold. 
(2.)J4.  .-.  60%— 540  acres, 

5.       1%=^.  of  540  acres=9  acres,  and 
ll.{  6.  100%=100   times   9    acres— 900   acres=what  he 

owned. 

(3.)  45%— what  he  owned. 

(4.)         .-.45%— 900  acres, 
(5.)  1%=^  of  900  acres— 20  acres,  and 

(6.)         100%— 100  times  20  acres— 2000=number  of  acres 

in  the  tract. 
III.     .-.  The  tract  contained  2000  acres. 

(R.  3d  p.,  p.  204,prob.  12.) 


MISCELLANEOUS  PROBLEMS. 


67 


I. 


!!.< 


A,  wishing  to  sell  a  cow  and  a  horse  to  B,  asked  150% 
more  for  the  horse  than  for  the  cow;  he  then  reduced 
the  price  of  the  cow  25%,  and  the  horse  33£%,  at  which 
price  B  took  them,  paying  $290;  what  was  'the  price  of 

each  ? 


1. 

2- 
(3. 

(4. 


100%—  asking  price  of  the  cow.     Then 
100%-H50%=250%=asking  price  of  the  horse. 
100%—  25%=75%=selling  price  of  cow. 
100%=250%, 

l%=nnr  of  250%=2.50%,  and 
33£%=33l  times  2.5%=83£%=reduction  on  the 

asking  price  of  the  horse. 

250%—  83i%=166f%=selling  price  of  the  horse. 
75%+166f%=241f%=  selling  price  of  both. 
$2QO=selling  price  of  both. 
/.  241f%=$290, 

°f    290==1-20   and 


)         75%—  75   times$  1.20=$90=selling  price  of   the 

cow. 
)       166S%=166f   times  $1.20=$200=selling  price  of 

the  horse. 


f$  90=selling  price  of  the  cow,  and 
''•|$200=selling  price  of  the  horse. 

(Brooks'  H.  A.,  p.  2J.3,  prob.  18.) 


I.  A  mechanic  contracts  to  supply  dressed  stone  for  a  church 
for  $87560,  if  the  rough  stone  cost  him  18  cents  a  cubic 
foot;  but  if  he  can  get  it  for  16  cents  a  cubic  foot,  he 
will  deduct  5%  from  his  bill;  required  the  number  of 
cubic  teet  and  the  charge  for  dressing  the  stone. 

100%=$87560. 

1%=T^  of  $87560=$875.60,  and 

5%=5  times  $875.60=$4378=the  deduction. 

18/  —  16/=2/—  the  deduction  per  cubic  foot. 

...  $4378=the  deduction  of  4378-7-.02,  or  218900  cubic 
feet.     Then 

$87560=cost  of  218900  cubic  feet. 

$.40=$87560-r-218900=:cost  of  one  cubic  feet. 

.-.  $.40  —  $.  18=$.  22=cost  .of  dressing  per  cubic  foot. 


Ill 


r218900=number  of  cubic  feet,  and 
'  '122  cents=cosc  of  dressing  per  cuoic  foot. 

(Brooks'  H.  A.,  p. 


,  prob.  21.) 


68  FINKEL'S   SOLUTION  BOOK. 


EXAMPLES. 

1.  A  merchant,  having  $1728   in  the  Union  Bank,  wishes  to 
withdraw  15%;  how  much  will  remain?  Ans.  $1468.80. 

2.  A  Colonel  whose  regiment  consisted  of  900  men,  lost  8% 
of  them  in  battle,  and  50%   of   the    remainder  by  sickness;  how 
many  had  he  left?  Ans.  414  men. 

3.  What  %  of  $150  is  25%  of  $36?  Ans.  6%. 

4.  What  %  of  I  of  f  off  is  -J?  Ans.  Bl±%. 

5.  If  a  man  owning  45%  of  a   mill,  should   sell  33-j-%   of  his 
share  for  $450  ;  what  would  be  the  value  of  the  mill  ? 

Ans.  $3000. 

6.  A.  expends   in   a  week   $24,  which   exceeds   by  33^%    his 
earnings  in  the  same  time.     What  were  his  earnings?     Ans.  $18. 

7.  Bought  a  carriage  for  $123.06,  which  was  16%  less  than  I 
paid  for  a  horse;  what  did  I  pay  for  the  horse?         Ans.  $146.50. 

8.  Bought  a  horse,  buggy,  and  harness   for  $500.     The  horse 
cost  37-^%    less  than  the   buggy,  and    the   harness   cost  70%  less 
than  the  horse  ;  what  was  the  price  of  each? 

Ans.  buggy  $275ff  ,  horse  $172^f,  and  harness  $51f£. 

9.  I  have  20  yards  of  yard-  wide  cloth,  which  will  shrink  on 
sponging  4%    in  length  and   5%  in  width;  how  much   less    than 
20  square  yards  will  there  be  after  sponging?        Ans.  l|~f  yards. 

10.  A.  found  $5;  what  was  his  gain  %?  Ans.  oo. 

11.  The  population  of  a  city  whose  gain  of  inhabitants   in  5 
years  has  been  25%,  is  87500  ;  what  was  it  5  years  ago? 

Ans.  70000. 

12.  The  square  root  of  2  is  what  %  of  the  square  root  of  3  ? 

Ans.    /~ 


13.  A  laborer  had   his  wages   twice  reduced  10%;   what  did 
he  receive  before  the   reduction,  if  he   now  receives  $2.02-^  per 
day?  Ans.  $2.50. 

14.  The  cube  root  of  2985984  is  what  %  of  the  square  root  of 
the  same  number?  Ans. 


15.  A  man  sold  two  horses  for  the  same  price  $210  ;  on  one  he 
gained  25%,  and  on  the  other  he  lost  25%;  how  much  did  he 
gain,  supposing  the  second  horse  cost  him  f  as  much  as  the  first? 

Ans.  $10. 


COMMISSION.  69 

16.  A  merchant  sold  goods  at  20%  gain,  but  had  it   cost  him 
$49  more  he  would   have  lost  15%  by  selling  at   the  same  price; 
what  did  the  goods  cost  him?  Ans.  $119. 

17.  If  an  article  had    cost    20%  more,  the   gain  would    have 
been  25%  less;  what  was  the  gain  %  ?  Ans.  50%. 


1        ^ 


II.     COMMISSION. 


1.  Commission  is  the  percentage  paid  to  an  agent  for  the 
transaction  of  business.     It  is  computed  on  the  actual  amount  of 
the  sale. 

2.  An  Agent,  Factor,  or  Commission  Merchant, 

is  a  person  who  transacts  business  for  another. 

3.  The  Net  Proceeds  is  the  sum  left  after  the  commission 
and  charges  have  been  deducted  from  the  amount  of  the  sales  or 
collections. 

4.  The  Entire  Cost  is  the   sum   obtained   by  adding  the 
commission  and  charges  to  the  amount  of  a  purchase. 

I.  An  agent  received  $210  with  which  to  buy  goods  ;  after 
deducting  his  commission  of  5%,  what  sum  must  he 
expend  ? 

1.  100%=what  he  must  expend. 

2.  5%=his  commission. 

3.  100%+5%=105%=what  he  receives. 
II. S  4.  $210=what  he  receives. 

6.  1  %=rhr  of  $210=$2,  and 

7.  100%=1C9  times  $2=$200=what  he  expends. 

III.   .-.  $200=what  he  must  expend. 

(R.  3d  p.,  p.  207,prob.  4.) 

Note. — Since  the  agent's  commission  is  in  the  $210,  we  must 
not  take  5%  of  $210;  for  we  w*ould  be  computing  commission  on 
his  commission.  Thus,  5%  of  ($200+$10)=$10+$.50.  This  is 
•$.50  to  much 

I,  An  agent  sold  my  corn,  and  after  reserving  his  com- 
mission, invested  the  proceeds  in  corn  at  the  same  price; 
his  commission,  buying  and  selling  was  3%,  and  his 
whole  charge  $12;  for  what  was  the  corn  first  sold? 


70 


FINKEL'S  SOLUTION  BOOK. 


II. 


in. 

i. 


a 

3.) 


(4.)- 


100%— cost  of  the  corn. 
3%— the  commission. 
100%—  3%— 97%— net  proceeds,  which  he  invested 

in  corn. 

1.   100%— cost  of  second  lot  of  corn, 
commission. 


entire   cost  of  second    lot  of- 


=co&t  of  second 


=          of  $12—  $2.06,  and 


3.  100%+3%— 103%: 

corn. 

4.  97%— entire  cost  of  second  lot  of  corn. 

5.  .-.  103%— 97%, 

6-       1  %=^  of  97%—  jfa% ,  and 

7.  100%— 100  times  •£$*%=&±£fo% 

lot  of  corn  in  terms  of  the  first. 

8.  3%— 3   times   Ty^%—  2T8-^%  —  com  mission    on 
second  lot. 

(5.  3%+2T8-053-%—5T8TF\%— whole  commission. 

(6.  $12— whole  commission. 

(7- 

(8.) 

(9. )   100%— 100  times  $2  06— $206— cost  of  first  lot  of  corn 
•.  $206— cost  of  first  lot  of  corn.   (R.H.  A., p.  219,prob.  10.) 

Sold  cotton  on  commission,  at  5%;  invested  the  net  pro- 
ceeds in  sugar,  commission,  2%;  my  whole  commission 
was  $210  ;  what  was  the  value  of  the  cotton  and  sugar? 
r  (1.  100%— cost  of  cotton. 

(2.  5%— commission. 

(3.          100%—  5%—  95%—  net 

1.  100%'— cost  of  sugar. 

2.  2  %— commission. 

3.  102%— entire  cost  of  sugar. 

4.  95%— entire  cost  of  sugar. 

5.  .-.  102%— 95%, 

6.  1%=^  of  95%—  T9Ti5Y%,  and 

7.  100%— 100    timesT9^%—  93-^%—  cost  of  sugar  in 

terms  of  cotton. 

8.  2%— 2  timesY^%  —1|4%— commission  on  the 
sugar. 

(4.)  5%4-l|T%— 6|4 %— whole  commission. 

(5.)         $210=\vhole  commission. 
(6.)        .-.  6|{%=$210, 

(7.)  1  %  =J4  of  $210=$30.60,  and 

(8.) 
(9.) 

TIT    •  J$3060— cost  of  cotton,  and 
'''^$2850— cost  of  sugar. 


[vested  in  sugar, 
proceeds,  which    he    in- 


100%— 100  times  $30.60— $3060— cost  of  cotton. 

— 93-^  times  $30.60— $2850— cost  of  sugar. 


.  H.  A., p.  219,prob.  6.) 


COMMISSION. 


71 


II. 


III 


III 
I 


A  lawyer  received  $11.25  for  collecting  a  debt ;  his  com- 
mission being  5%;  what  was  the  amountof  the  debt? 

1.  100%=amount  of  the  debt. 

2.  5%=commission. 
4.   $11.25=commission. 

4.  .-.  5%=$11.25. 

5.  l%=i  of  $11.25=$2.25,  and 

6.  100%— 100  times  $2.25=$225=an)outtt  Of  the  debt. 
.'.  $225=amount  of  debt. 

(7?,  3d  p.,  p.  207,  prob.  6.) 

Charge  $52.50  for  collecting  a  debt  of  $525;  what  was  the 

rate  of  commission? 
1.  $525—100% 
!!.<!  2.  $1=^  of  100%—^-%,  and. 

3.  $52.50=52.5  times  ^-%=10%=rate  of  commission. 
.'.  10%=rate  of  commission. 

My  agent  sold  my  flour  at  4%  commission;  increasing  the 
proceeds  by  $4.20,  I  ordered,  the  purchase  of  wheat  at 
2%  commission;  after  which,  wheat  declining  3£%, 
my  whole  loss  was  $5  ;  what  was  the  flour  worth? 

100%—cost  of  flour. 

4%=commission  on  flour. 
100%—  4%=96%=net  proceeds. 

1.  100%=cost  of  wheat. 

2.  2%=commission  on  wheat. 

3.  100%+2%=102%=entire  cost  of  wheat. 

4.  96%+$4.20=entire  cost  of  wheat. 

5.  .-.  102 %=96% +$4.20, 

6.  i%_T^_0f  (96%+$4.20)= .94T2T%+$.0411|f, 

7.  100%=100  times  (.94T2T%+$.0411|f  )=94&%+ 

$4.11|f=cost  of  wheat. 

8.  2%=2  times  (.94T27%+$.0411i-f-)=ll|-%+$.08TV 
II  ^             I          —commission  on  wheat. 


[il 

(3.) 


(4.) 


5=3^-  times 
loss  on  wheat. 

4%  +  Hf  %  +  $-08T4T  +  3-gV%  +  $• 
$.21ff=whole  loss. 
$5=whole  loss. 

9JT%=^$5— $.21|f— $4-78^2T. 
1%=J-  of  $4.178^r=$.53,  and 


III. 


(6.) 

(7-) 
(8.) 
(9.) 

(10.) 

(11.)         100%=^100  times  $.53=$53. 
.-.  $53=cost  of  flour.  ( /?.  H.  A., p.  219,  prob. 


72  FINKEL'S   SOLUTION  BOOK. 


EXAMPLES. 

1.  A  broker  in  New  York  exchanged  $4056  on  Canal    Bank, 
Portland,  at  -f- %;  what  did  he  receive  for  his  trouble? 

Ans.  $26.35. 

2.  A  sold  on  commission  for  B  230  yards  of  cloth  at  $1.25  per 
yard,  for  which  he  received  a  commission  of  3^%;   what  was  his 
commission  and  what  sum  did  he  remit? 

Ans.  Commission  $10.06^,  and  Remittance  $277.43}. 

3.  A  sold  a  lot  of  books  on  commission  of  20%,  and  remitted 
$160;  for  what  were  the  books  sold?  Ans.  $200. 

4.  A  lawyer  charged  $80  for  collecting  $200;    what  was  his 
rate  of  commission?  Ans.  40% 

5.  I  sent  my  agent  $1364.76  to  be  invested  in  pork  at  $6  per 
bbl.  after  deducting  his  commission  of  2%;  how  many  barrels  of 
pork  did  he  buy?  Ans.  223  bbl. 

6.  How  much  money  must  I  send  my  agent,  so  that  he  may 
purchase  250  bbl.  of  flour  for  me  at  $6.25  per  bbl.,  if  I  pay  bim 
2%%  commission?  Ans.  $1601.5625. 

7.  If  an  agent's  commission  was  $200,  and  his  rate  of  com- 
mission 5%  ;  what  amount  did  he  invest?  Ans.  $4000 

8.  My   agent  sold  cattle  at  10%   commission,  and  after  I  in- 
creased the  proceeds  by  $18,  I  ordered  him  to  buy  hogs  at  20% 
commission.     The  hogs  had  declined  6f%,  when  he  sold  them  at 
14f%  commission.     I  lost  in  all  $86;  whatdidthe  cattle  sell  for? 

Ans.  $200. 

9.  An  agent  sells  flour  on  commission  of  2%,  and  purchases 
goods  on  true  commission  of  3%.       If  he  had  received  3%  for 
selling  and  2%   for  buying,  his  whole  commission   would   have 
been  $5  more.       Find  the  value  of  the  goods  bought. 

Ans. 

III.     TRADE  DISCOUNT. 

1.  Trade    Discount  is  the  discount  allowed  in  the  pur- 
chas  and  sale  of  merchandise. 

2.  A  List,  or  Hegular  JPrice,  is  an  established  price,  as- 
sumed by  the  seller  as  a  basis  upon  which  to  calculate  discount. 

3.  A  Ifet  Price  is  a  fixed  price  from  which  no  discount 
is  allowed. 

4.  The  Discount  is  the  deduction  from  the  list,  or  regu- 
.lar  price. 


TRADE  DISCOUNT. 


73 


II. 


III. 


Sold  20  doz.  feather  dusters,  giving  the  purchaser  a  dis- 
count of  10,  10  and  10%  off,  his  discounts  amounting  to 
$325.20;  how  much  was  my  price  per  dozen? 

(1.)         100%— whosesale  price. 

(.2.1  10%  of  100%— 10%— first  discount. 

(3.)         100%— 10%— 90%— first  net  proceeds. 

100%— 90%, 

1%— Tio  of  90%— &%,  and 
10%— 10  times  T9^%z=9%— second  discount. 
90^— 9 %— 81%— second  net  proceeds, 
rl.  100%— 81%, 

\3.     10%— 10°times  ^%==8.i%==third  discount. 
(6.)         10%+9%+8.1%—  27.1%— sum  of  discounts. 
(7.)         $325.20— sum  of  discounts. 
(8.)        .-.  27.1%— $325.20, 
(9.)         1%— ^T  of  $325.20— $12,  and 
(10.)         100%— 100    times    $12— $1200— wholesale    price 

of  20  dozen. 
(11.)         $60— $1200-5-20= wholesale  price  of  1  dozen. 

.•.  $60— wholesale  price  per  dozen. 

(.ff.  3d  p.,  p.  209,  prob.  5.) 

Bought  100  dozen  stay  bindings  at  60  cents  per  dozen  for 
40,  10,  and  7^%  off;  what  did  I  pay  for  them? 


(I-) 
(2.) 


(3.)- 


(4.)' 


(5.). 


60/=  list  price  of  1  dozen. 
$60—100  times  $.60— list  price  of  100  dozen. 

1.  100%— $60, 

2.  1  %—  y^-  of$60— $.60,  and 

3.  40%— 40  times  $.60— $24=first  discount. 

4.  $60— $24— $36— first  net  proceeds. 

1.  100%— $36, 

2.  1  %—  y-^j-  of  $36— $.36,  and 

3.  10%— 10  times  $.36— $3.60=second   discount. 
$36 — $3»60— $32.40— second  net  proceeds. 

100%— $32.40, 

1%=.^.  of  $32.40=$.324,  and 
7|%— 7|  times  $.324— $2.43— third  discount. 


2. 
3. 
,4.  $32.40— $2.43— $29.97— cost. 


I  paid  $29.97. 


3d  p.,  p.  209,  prob.  6.) 


I.  A  retail  dealer  buys  a  case  of  slates  containing  10  dozen 
for  $50  list,  and  gets  50,  10,  and  10%  off;  paying  for 
them  in  the  usual  time,  he  gets  an  additional  2%;  what 
did  he  pay  per  dozen  for  the  slates? 


74  FTNKEL'S   SOLUTION  BOOK. 

100%— $50. 

2.  1  %— Ho  of  $50— $.50. 

3.  50%— 50  times  $50— $25— first  discount. 
A.  $50— $25— $25— first  net  proceeds. 

100%— $25. 

l^=T^rof$25=$.25. 

10%=10  times  $.25— $2.50— second  discount 
._    $25— $2.50— $22.50— second  net  proceeds, 
rl.  100%— $22.50. 
1 2.       1%—TTnr  of  $22.50=4.225. 

*' n  s      10%— 10  times  $.225=$2.25=third  discount. 
$22.50— $2.25— $20.25=third  net  proceeds. 
100%=-$20.25. 

1%=^  of  $20.25=$.2025. 
2%=2  times  $.2025=$.405=fourth  discount. 
$20.25— $.405=$19.845=cost  of  10  dozen  slates. 
$1.9845=$19.845-f-10=:cost  of  1  dozen  slates. 

III.     .'.  $1.9845=cost  of  1  dozen  slates. 

(R.  3d  p.,  p.  209,prob.  9.) 

I.  Sold  a  case  of  hats  containing  3  dozen,  on  which  I  had  re- 
ceived a  discount  of  10%  and  made  a  profit  of  12^%  or 
37-JX  on  each  hat ;  what  was  the  wholesale  merchant's 
price  per  case? 

(1.)  37!/==profit  on  one  hat. 

(2.)  $13.50—36  times  $.37|—profit  on  3  dozen  hats. 

(3.)  100%— wholesale  merchant's  price  per  case. 

(4.)      10%=discount. 

(5.)  100%— 10%— 90%—my  cost. 


(6.)  12.       1%=^  of  90%— .9%. 

]3.  12|%— 12|  times  .9%— lli%=profit  in  terms  of 
wholesale  price. 

(7.)         .'.  lli%—  $13.50'. 

(8.)         l%==*iTT  of  $13.50— $1.20. 


(9.)         100%— 100    times    $1.20=$120=wholesale- mer- 
chant's price  per  case. 

HI.     .'.  $120=wholesale  merchant's'price  per  case. 

(7?.  3d  p.,  p.  212,prob.  4.) 

I.  A  bookseller  purchased  books  from  the  publishers  at  20% 
off  the  list ;  if  he  retail  them  at  the  list  what  will  be 
his  per  cent,  of  profit? 


TRADE  DISCOUNT. 


II. 


1.  100%=list  price. 

2.  20%=discount. 

3.  100%—  20%=80%=cost. 

4.  100%=bookseller's  selling  price,  because  he  sold  them 

at  the  list  price. 

5.  .-.  100%—  80%=20%=gain. 

6.  80%=100%  of  itself. 

7.  1%=-^  of  100%— li%,  and 

8.  20%=20  times  l^%=25%=his  gain  %. 


(R.  3d  p.,  p.  211,prob.  1.) 


III.  .-.  25%=his  %  of  profit. 


Note. — Observe  that  since  his  cost  is  80%,  and  his  gain 
we  wish  to  know  what  %  20%  is  of  80%.  It  will  become  evi- 
dent if  we  suppose  the  list  price  to  be  (say)  $400,  and  then  pro- 
ceed to  find  the  %  of  gain  as  in  the  above  solution. 

Bought  50  gross  of  rubber  buttons  for  25,  10,  and  5%  off ; 
disposed  of  the  lot  for  $35.91,  at  a  profit  of  12%  ;  what 
was  the  list  price  of  the  buttons  per  gross? 

(1.)         100%=Hst  price. 
(2.)  25%  of  !00%=25%=first  discount. 

100%—  25%=75%=first  net  proceeds. 
100%=75%, 

l%=yi¥  of  75%=|%,  and 
10%=10  times  j%=7£%. 
75%— 7i%=67|%= second  net  proceeds 


ir 


1. 

2. 

3. 

,4. 
1. 

2.  1%=T^  of  67i%=.67£%,  and 

3.  5%=5  times  .67i%=3.375%=third  discount 

4.  67t%— 3.375%=64.125%=cost. 

1.  100%=64.125%, 

2.  !%=.64125%,and 

3.  12%=12  times  ,64125 %=7.695%=gain. 

4.  .-.  64.125%-f  7.695%=71.82%=selling  price. 
$35.91=selling  price. 

...  71.82%=$35.91, 

1%=^^  of  $35.91=$.50,  and 

100%  =100 'times  $.50=$50=list  price  of  50  gross, 

$1.00=$50-;-50=list  price  of  one  gross. 

til.  .'.  $1.00=list  price  of  one  gross. 

(R.  3d  p.,  p.  212,prob.  10.) 

I  A  dealer  in  notions  buys  60  gross  shoestrings  at  70/  per 
gross,  list,  50,,  10,  and  5%  off;  if  he  sell  them  at  2d 
10,  and  5%  off  list,  what  will  be  his  profit? 


76 


II. 


FINKEL'S   SOLUTION  BOOK. 

70/=list  price  of  one  gross. 

$42=60  times  $.70=list  price  of  60  gross. 

100%=$42. 


50  times  $.42=$21s=first  discount. 
$42—  $21=$21=first  net  proceeds. 


10$F==10  times  $.21=$2.10=second  discount. 
$21—  $2.10=$18.90=  second  net  proceeds. 

100%=$18.90. 


(7.) 


5%=$.945=third  discount 
$18.90— $.945=$!  7.955=cost 
100%=$42 

1  %=T£Tr  of  $42=$.42.  [count. 

20%=20  times  $.42=$8.40=first  conditional  dis- 
4.  $42— $8  40=$33.60=first  conditional  net  proceeds. 

1.  100%=$33.60. 

2.  1  %=^  of  $33.60=$.336.  [discount 

3.  10%=10  times  $.336=$3.36=second  conditional 

4-   $33.60— $3.36=$30.24= second   conditional   net 
proceeds. 

1.  100%=$30.24. 

2.  1  %=rf(r  of  $30.24=$.3024.  [discount 

3.  5%=5  times  $.3024=$1.512=third  conditional 

.24— $1.512=$28.728=selling  price. 
$28.728— $17.955=$10.773=his  profit. 


III.  .'.  $10.773=his  profit 


(R.  3d  p.,  p.  212,  prob.  9.) 


EXAMPLES. 

1.  Bought  a  case  of  slates  containing  12  doz.  for  $80  list,  and 
got  45,  10,  and  10%  off;  getting  an  additional  2%  off  for  prompt 
payment,  what  did  I  pay  per  dozen  for  the  slates? 

Ans.  $2.9106. 

2.  Bought  a  case  of  hats   containing  4  doz.,  on  which   I  re- 
ceived a  discount  of  40.  20,  10,  5,  and  2£%  off.     If  I  sell  them  at 
$4  a  piece  making  a  profit  of  20%,  what   is   the  wholesale  mer- 
chant's price  per  case?  Ans.  $399|Jf§^. 

3.  If  I  receive  a  discount  of  20,  10,  and  5%  off,  and  sell  at  a 
discount  of  10,  5,  and  2^%  off;  what  is  my  %  of  gain? 

Ans.  21|%— . 

4.  A   bill    of    goods   amounted    to    $2400;    20%    off    being 
allowed,  what  was  paid  for  the  goods?  Ans.  $1920. 

5.  -  Bought  goods  at   25,  20,  15,  and  10%  off.     If  the  sum  of 
my  discounts  amounted  to  $162.30,  what  was  the  list  price  of  the 
goods?  '  Ans.  $300 


PROFIT  AND  LOSS.  77 


IV.     PROFIT  AND  LOSS. 

1.     Profit  and  Loss  are  terms  which  denote  the  gain  or  loss 
in  business  transactions. 


2. 
3. 

I. 


III. 


I. 


II. 


Profit  is  the  excess  of  the  selling  price  above  the  cost. 
LOSS  is  the  excess  of  the  cost  above  the  selling  price. 

A  merchant  reduced  the  price  of  a  certain  piece  of  cloth 
5  cents  per  yard,  and  thereby   reduced   his   profit  on  the 
cloth  from  10%  to  "8%;    what  was  the  cost  of  the  cloth 
per  yard? 
1.   100%=cost  of  cloth  per  yard. 

10%=his  profit  before  reduction. 

8%=his  profit  after  reduction. 
10%—  8%=2%=his  reduction. 
5/—  reduction. 


=2        and 


100%  =100  times  2^/=$2.50=cost  per  yard. 
.-.  $2.50=cost  of  cloth  per  yard. 

(R<  3d  p.,  p.  211,prob.  13.) 

A  dealer  sold  two  horses  for  $150  each;  on  one  he  gained 
25%  and  on  the  other  he  lost  25%;  how  much  did  he 
lose  in  the  transaction  ? 

1.)         100%=cost  of  the  first  horse; 

2.)  25%=gain. 

3.)         100%+25%=125%=selling  price  of  first  horse. 

4.)         $150—  selling  price. 
(5.)         .'.  125%=$150, 
(6.)  l%=Tiir  of  $150=11.20,  and 

(7.)         100%=100  times  $1.20=$120=cost  of  first  horse. 
(8.)         $150—  $120=$30=gain  on  first  horse. 

1.  100%=cost  of  second  horse. 

2.  25%=loss. 

3.  100%  —  25%=75%=selling  price  of  second  horse. 


(9.) 


4.   $150=selling  price. 


III. 


I. 


5.  .-.  75%=$150, 

6.  1%=  TV  of  $150=$2,  and 

7.  100%=400  times  $2=$200=cost  of  second  horse. 
$200  —  $150=$50=loss  on  second  horse. 

$50  —  $30=$20=loss  in  the  transaction. 

He  lost  $20  in  the  transaction. 

(  R.  3d  p.  ,  p.  211,  prod.  12.  ) 

A  speculator  in  real  estate  sold  a  house  and  lot  for  $12000, 
which  sale  afford  him  a  profit  of  33^%  on  the  cost  ;    he 


(11.) 


78 


FINKEL'S   SOLUTION  BOOK. 


then  invested  the  $12000  in  city  lots,  which  he  was 
obliged  to  sell  at  a  loss  of  33ij%;  how  much  did  he  lose 
by  the  two  transactions? 


II. 


(!•) 
(2.) 
(3.) 

(4.) 
(5.) 

(6.) 


100%—  cost  of  the  house  and  lot. 

33^%—  gain.  [lot. 

100%+33£%=133i%=selling  price  of  house  and 

$12000—  selling  price  of  the  house  and  lot. 

.-.  133i%=$12000. 


(7.) 
(8.) 

(9.) 
(10.) 


=^1  of  $12000=190. 
100$ 


[lot. 


III. 
I. 


II. 


100%— 100  times  $90— $9000— cost  of  house  and 
$12000— $9000— $3000— gain  on  house  and  lot. 
100%— $12000. 

1%— -^  of  $12000— $120. 

33^%—  33J  times  $120— $4000— loss  on  city  lots. 
$4000— $3000— $1000— loss  by   the    two    transac- 
tions. 
.\  $1000— his  loss  by  the  two  transactions. 

(./?.  3d  p.,  p.  211,  prob.  15. ) 

A   dealer   sold   two    horses  for  the  same  price;    on  one  he 
gained  20%,  and   on  the  other  he   lost  20%;  his  whole 
loss  was  $25  ;  what  did  each  horse  cost? 
(1.)         100%— selling  price  of  each  horse, 
'l.   100%— cost  of  first  horse. 

20%— gain  on  the  first  horse. 
100%+20%— 120%— selling  price  of  first  horse. 
...  120%— 100%,  from  (1), 

1%— T-b  of  100%—f%,  and 
100%— 100  times  |%=83^%— cost  of  first  horse 

in  terms  of  the  selling  price. 
100%—  83i%— 16f%—  gain  on  first  horse. 

100%— cost  of  the  second  horse. 

20%— loss  on  second  horse. 

100%—  20%— 80%—  spelling  price  of  second  horse. 
.-.80%— 100%,  from  (1), 

1%— ¥V  of  100%— li%,  and 
100%— 100    times    1|%=125%—  cost    of    second 

horse  in  terms  of  the  selling  price. 
125% — 100%— 25%— loss  on  the  second  horse. 
25%—  16f  %— 8J%— whole  loss. 
$25— whole  loss. 


(7.) 

(8.) 

(9.) 

((10.) 


1. 

2. 

3. 

4. 

(z.  )< 

5. 

6. 

7. 

i. 

2. 

3! 

(3.)< 

4. 

5. 

6. 

7. 

(4.) 

(5.) 

(6.) 

l%:=_of$25=$3,  and 

o^  [horse. 

100%=100  times  $3=$300=selling  price  of  each 
83£%=83£  times  $3=$250= cost  of  first  horse. 
125%=125  times  $3=$375=cost  of'second  horse. 


PROFIT  AND  LOSS. 


79 


.  r$250=cost  of  the  first  horse,  and 
|  $375=cost  of  second  horse. 

I.     What  %  is  lost  if  f  of  cost  equals  f  of  selling  price  ? 

1.  f  of  selling  price=|-  of  cost. 

2.  £  of  selling  price=£  of  f  of  cost=f  of  cost. 

3.  J-  of  selling  price=4  times  -J  of  cost=f  of  cost. 

4.  |=cost. 

5!  !=selling  price. 

B.  J=fr  of  100%=1H%,  loss. 
III.  .-.  Loss=lli%. 


II. 


I. 


II. 


Paid  $125  for  a  horse,  and  traded  him  for  another,  giving 
60%  additional  money.  For  the  second  horse  I  received 
a  third  and  $25.  I  then  sold  the  third  horse  for  $150 ; 
what  was  my  %  of  profit  or  loss? 


(2.) 
(3.) 

(4.) 


(7.) 
(8. 


100%=$125, 

1%==^  of  $125=$1.25,  and 
60%=60    times   $1.25=$75  =  additional   money 
paid  for  the  second  horse. 
$125+$75=$200=cost  of  second  horse. 
$150=selling  price  of  the  third  horse. 
$150+$25=$175=selling  price  of  second  horse. 
$200— $175=$25=loss  in  the  transaction. 
$200=100%, 

$l=innr  of  100%=!%,  ana 
$25=25  times  |%=12|%=my  loss. 


III.  .-.  My  loss  is  12|%. 


(./?.  H.  A.,  p.  201,prob.  4.) 


I.     If  [  buy  at  $4  arid  sell  at  $1,  how  many  %  do  I  lose? 

'1.  $4=cost 

2.  $l=selling  price. 

3.  $4— $l=$3=loss. 

4.  $4=100%. 

5.  $l=i  of  100%=25%. 

6.  $3=3  times  25%=75%=loss. 

III.  .-.  75%=loss. 

I.     A  and  B  each  lost  $5,  which  was  2£%  of  A's  and 
B's  money  ;  which  had  the  most,  and  how  much? 


of 


80 


FINKEL'S  SOLUTION  BOOK. 


II. 


III. 


(1.)  100%=A's  money. 

(2.)  2J%— whathe  lost. 

(3.)  $5— what  he  lost. 

(4.)  .-.  2£%— $5, 

(5.)  1%=^7  of  $5— $1.80,  and 

(6.)  100%— 100  times  $1.80— $180=A's  money. 

1.  100%— B's  money. 

2.  3£%—  what  he  lost. 

3.  $5— what  he  lost. 

5.  1  %=4-  of  $5=$1.50,  and 

3-$ 

6.  100%— 100  times  $1.50=$150=B's  money. 

(8.)  $180— $150— $30— excess  of  A's  money  over  B's. 
.%  A  had  $30  more  than  B.          (R.  H.  A.,  p.  203,  prob,  5.) 


I.  Mr.  A  bought  a  horse  and  carriage,  paying  twice  as  much 
for  the  horse  as  for  the  carriage.  He  afterward  sold  the 
horse  for  25%  more  than  he  gave  for  it,  and  the  carriage 
for  20%  less  than  he  gave  for  it,  receiving  $577-50;  what 
was  the  cost  of  each? 


I. 


100%—  cost  of  the  carriage. 
200%=cost  of  the  horse. 

20%=loss  on  the  carriage. 

100%—  20%=80%=selling  price  of  the  carriage. 
100%—  200%, 

l%=TTnr  of  200%=2%,  and 

25%  =25  times  2%=50%=  gain  on  the  horse. 
200%+50%=250%=selling  price  of  the  horse. 

80%+250%=330%=selling  price  of  both. 
$577.  50=selling  price  of  both. 
.-.  330%=$577.50, 

l%=irb  of  $577.50—  $1.75,  and 
100%=1  00  times  $1.75=$175=cost  of  carnage. 
200%=200  times  $1.75—  $350=cost  of  the  horse. 

>  J$175=cost  of  the  carriage,  and 
•'•  |$350—  cost  of  the  horse. 

(Milne's  prac.,  p.  259,  prob.  19.  ) 

Mr.  A.  sold  a  horse  for  $198,  which  was  10%  less  than  he 
asked  for  him,  and  his  asking  price  was  10%  more  than 
the  horse  cost  him.  What  did  the  horse  cost  him? 


II. 


II. 


PROFIT  AND  LOSS.  81 

(1.)  100%— cost  of  the  horse. 

(2.)  100%  +  10%— 110%— asking  price. 

rl.  100%— 110%, 
(8..K2.        l%=Tio  of  H0%— 1TV%,  and      [asking  price. 

13.     10%— 10    times    lT17r%=ll%  =  reduction    from 

(4.)  110%— 11%—  99%=selling  price. 

(5.)  $198— selling  price. 

(6.)  .-.  99%— $198, 

(7.)  1%— gV  of  $198— $2,  and  ' 

(8.)  100%— 100  times  $2— $200— cost  of  the  horse. 

III.  .-.  $200— cost  of  horse.         (Milne's  prac.,  p.  259,  prob.  23,) 

I.  What  must  be  asked  for  apples  which  cost  me  $3  per  bbln 
that  I  may  reduce  my  asking  price  20%  and  still  gain 
20%  on  the  cost? 

(1.)         100%— $3. 
(2.)  l%=rfo  of  $3— $.03,  and 

(3.)  20%— 20  times  $.03— $.60— gain. 

(4.)   $3.00+$.60— $3.60— selling  price. 
'1.   100%— asking  price. 

2.  20%— reduction. 

3.  100%—  20%— 80%— selling  price. 

4.  $3.60— selling  price. 

5.  .-.  80%— $3.60, 

6-       l%=8i>  of  $3-60— $.045,  and 

7.  100%— 100  times  $.045— $4.50— asking  price. 

ill.  .'.  $4.50— asking  price.  (Milne's prac., p.  261, prob.38.) 


(5.) 


I.     A  merchant  sold   a   quantity  of  goods  at  a  gain   of 

If,  however,  he  had  purchased  them  for  $60  less  than 
he  did,  his  gain  would  have  been  25%.  What  did  the 
goods  cost  him  ? 

100%— actual  cost  of  goods. 

20%— gain. 

100%+20%— 120%—  actual  selling  price. 
100% — $60— supposed  cost. 
'1.  100%=100%— $60, 

2.       1%=T^  of  (100%— $60)— 1%— $.60,  and 
!!.<!    ^y>']3.     25%— 25  times  (1%— $.60)— 25% —$15  =  sup- 
posed gain.  [ual  selling  price. 
(100%— $60)  +  (25%—  $15)— 125%—  $75— act- 
.•;  125%— $75— 120%, 
5%=$75, 

l%=|of  $75— $15,  and 
100%— 100  times  $15— $1500— cost  of  the  goods. 

III.   .-.  $1500— cost  of  goods.       (Milne's  prac.,  p.  261,  prob.  40.) 


82  FINKEL'S   SOLUTION   BOOK. 

Note, — The  selling  price  is  the  same  in  the  last  condition  of 
this  problem  as  in  the  first.  Hence  we  have  the  selling  price  in 
the  last  condition  equal  to  the  selling  price  in  the  first  as  shown 
in  step  (7.) 

I.     I  sold  an  article  at  20%  gain,  had   it  cost  me  $300  more,  I 
would  have  lost  20%;  find  the  cost. 


II. 


(1- 
(2. 

(3. 


100%— actual  cost  of  the  article. 

20%— actual  gain. 
100%+20%— 120%— actual  selling  price. 


(4.)         100% +$300— supposed  cost. 
1.  100%— 100%+$300, 

of  (100%+$300)— 1%+$3>  and 


!!.<!    vt"']3,     20%—  20 times  (1%-f  $3)— 20% +$60— supposed 
loss.  [ual  selling  price. 

(100% +$300)— (20% +$60)—  80%+$240=act- 
.-.  120%— 80%+$240. 
40%— $240, 

1%— ^o-  of  $240— $6,  and 
100%— 100  times  $6— $600— cost  of  the  article. 

III.  .-.  -$600— cost  of  the  article. 

(7?.  H.  A.,  p.  409,prob.  85.) 

I.  A  man  wishing  to  sell  a  horse  and  a  cow,  asked  three 
times  as  much  for  the  horse  as  for  the  cow,  but,  finding 
no  purchaser,  he  reduced  the  price  of  the  horse  20%, 
and  the  price  of  the  cow  10%,  and  sold  them  for  $165. 
What  did  he  get  for  each? 

1.)  100%— asking  price  of  the  cow. 

2.)  300%— asking  price  of  the  horse. 
3.)  10%— reduction  on  the  price  of  the  cow. 

>)  100% — 10%— 90%— selling  price  of  the  cow. 


(1.  100%— 300%, 
(5.K2.       l%=iio  of  300%— 3%,  and 

13.  20%—  20  times  3%—  60%— reduction  on  horse. 

(6.)  300%— 60%— 240%— selling  price  ot  the  horse. 

(7.)  90% +240%—  330%— selling  price  of  both. 

(8.)  $165— selling  price  of  both 

(9.)  .-.  330%— $165, 
(10.)  l%=idny  of  $165— $.50,  and 

(11.)  90%— 90  times  $.50— $45— selling  price  of  cow. 

(12.)  240%— 240    times    $.50— $120— selling    price    of 
horse. 

Ill    •  / $45— amount  he  received  for  the  cow,  and 
'    I  $120— amount  he  received  for  the  horse. 


PROFIT  AND  LOSS. 


EXAMPLES. 

1.  What  price  must  a  man  ask  fora  horse  that  cost  him  $200, 
that  he  may  iali  20%  on  his  asking  price  and  still  gain  20%  ? 

Ans.  $300. 

2.  A  man  paid  $150  for  a  horse  which  he  offered  in  trade  at 
a  price  he   was  willing  to  discount  at  40%  for  cash,  as  he  would 
then  gain  20%.     What  was  his^trading  price?  Ans.  $300. 

3.  A  man  gained  20%  by  selling  his  house  for  $3600.     What 
did  it  cost  him  ?  Ans.  $3000. 

4.  A  gained  120%  by  selling  sugar  at  8?  per  pound.       What 
did  the  sugar  cost  him  per  pound?  Ans.  3T7T/. 

5  How  must  cloth,  costing  $3.50  a  yard,  be  marked  that  a 
merchant  mavr  deduct  15%  from  the  marked  price  and  still  gain 
15%?  Ans.W&fr. 

6.  Sold   a  piece  of  carpeting  for  $240,  and  lost  20%;    what 
selling  price  would  have  given  me  a  gain  of  20%  ? 

Ans.  $360. 

7.  Sold   two  carriages   for   $240    apiece,  and  gained  20%  on 
one  and  lo?t  20%  on  the  other;    how  much  did  I  gain  or  lose  in 
the  transaction  ?  Ans.     Lost  $20. 

8.  Sold  goods  at  a  gain  of  25%   and  investing  the  proceeds, 
sold  at  a  loss  of  25%  ;  what  was  my  %  of  gain  or  loss.    Ans.  6J%. 

9.  A  man  sold  a  horse  and  carriage  for  $597,  gaining  by  the 
sale,  25%  on  the  horse  and  10%  on  the  cost  of  the  carriage.       If 
j  of  the  cost  of  the   horse   equals  f  of  the   the   cost  of  carriage, 
what  was  the  cost  of  each?         Ans.     Carriage  $270;  horse  $240. 

10.  If  ^  of  the  selling  price  is  gain,  what  is  the  profit? 

Ans.  80%. 

11.  If  -J-  of  an  article  be  sold   for  the  cost  of  -J  of  it,  what  is 
the  rate  of  loss?  Ans.  33^%. 

12..  I  sold  two  houses  for  the  same  sum;  on  one  I  gained  25% 
and  on  the  other  I  lost  25%.  My  whole  loss  was  $240;  what 
did  each  house  cost?  Ans.  First  $1440,  second  $2400. 

13.  My  tailor  informs  me  that  it  will  take  10i  sq.  yd.  of 
cloth  to  make  me  a  full  suit  of  clothes.  The  cloth  I  am  about  to 
buy  is  1J  yards  wide  and  on  sponging  it  will  shrink  5%  in  length 
and  width.  How  many  yards  will  it  take  for  my  new  suit? 

Ans. 


14.     A  grocer  buys  coffee  at  15/  per  ft>.  to  the  amount  of  $90 
worth,  and  sells  it  at  the  same  price  by  Troy  weight  ;  find  the 
of  gain  or  loss.  Ans.    Gain 


84  FINKEL'S   SOLUTION  BOOK. 

15.  I  spent  $260  for  apples  at  $1.30  per  bushel  ;  after  retain- 
ing a  part  for  my  own  use,  I  sold  the  rest  at  a  profit  of  40%, 
clearing$13on  the  whole  cost.  How  many  bushels  did  I  retain? 

Ans.  50  bu. 

16  How  must  cloth  costing  $3.50  per  yard,  be  marked  that 
the  merchant  may  deduct  15%  from  the  marked  price  and  still 
make  15%  profit?  Ans.  $4.735. 

17.  1  sold  goods  at  a  gain  of  20%.     If  they  had  cost  me  $250 
more  than  they  did,  I  would  have  lost  20%  by  the  sale-       How 
much  did  the  goods  cost  me?  Ans.  $500. 

18.  A  merchant   bought    cloth  at  $3.25  per  yard,  and  after 
keeping  it  6  months  sold  it  at  $3.75  per  yard.      What    was    his 
gain  %,  reckoning  6%  per  annum  for  the  use  of  money? 

Ans.  12%+. 


V.     STOCKS  AND  BONDS. 


1.  Stocks  is  a  general  term  applied  to  bonds,  state  and 
national,  and  to  certificates  of  stocks  belong  to  corporations. 

3.  A  Bond  is  a  written  or  printed  obligation,  under  seal, 
securing  the  payment  of  a  certain  sum  of  money  at  or  before  a 
specified  time. 

3.  Stock  is  the  capital  of  the  corporation  invested  in  busi- 
ness; and  is  divided  into  Shares,  usually  of  $100  each. 

4.  An  Assessment   is  a   sum   of   money  required   of  the 
stockholders  in  proportion  to  their  amount  of  stock. 

5.  A  Dividend  is  a  sum  of  money  to  be  paid  to  the  stock- 
holders in  proportion  to  their  amounts  of  stock. 

6.  The  far    Value  of  money,   stocks,  drafts,  etc.,  is  the 
nominal  value  on  their  face, 

7.  The  Market  Value  is  the  sum  for  which  they  sell. 

8.  Discount    is  the  excess    of   the    par    value    of   money, 
stocks,  drafts,  etc.,  over  their  market  value. 

9.  Premium  is  the  excess  of  their  market  value  over  their 
par  value. 

1C.  Brokerage  is  the  sum  paid  an  agent  for  buying  stocks, 
bonds,  etc. 


STOCKS  AND  BONDS. 


85 


II. 


I.     At  |%  brokerage,  a  broker  received  $10  for  making. an  in- 
vestment in  bank  stock  ;    how  many  shares  did  he  buy? 

1.  100%— par  value  of  stock. 

2.  i%— brokerage. 

3.  $10— brokerage. 

4.  .-.  i%=$10, 

5.  1%—  4  times  $10—  $40,  and 

6.  100%— 100  times  $40— $4000— par  value  of  stock. 

7.  $100— par  value  of  one  share. 

8.  $4000— par  value  of  4000-r-lOO,  or  40  shares. 

III.    •.  40— number  of  shares. 

I.     How  many  shares  of  railroad  stock  at  4%    premium  can 
be  bought  for  $9360? 

rl.   100%— par  value  of  stock  I  can  buy. 

2.  4%— premium. 

3.  104%— price  of  what  I  buy. 

4.  $9360— price  of  what  I  buy. 
II.<!5.  .-.  104%— $9360. 

6.  1  %=rht  of  $9360— $90. 

7.  100%— 100  times  $90— $9000— par  value. 
$100— par  value  of  one  share. 

^9.  $9000— par  value  of  9000-^-100,  or  90  shares. 

III.   .-.  90— number  of  shares  that  can  be  bought. 

I.     When  gold  is  at  105,  what  is  the  value  of  a  gold  dollar  in 
currency  ? 

1.  105X  ;  or  105%  in  currency— 100/;  or  100%  in  gold. 

2.  I/;  or  1%  in  currency— .95-^y/ ;  or  .95-^%  in  gold. 

3.  100X;  or  100%  in  currency— 95^TX 5  or  95^%   in  gold. 
.*.  $1  in  currency  is  worth  95^X  in  gold. 

In   1864,  the  "greenback"  dollar  was   worth  only  35f/  in 
gold;   what  was  the  price  of  gold  ? 

1.  35fX;  or  35f%  in  gold— 100/ ;  or  100%  in  currency. 

2.  I/;  or  1%  in  gold— -^  of  100X;  or   100%—  2.8X ;  or 
2.8%  in  currency.         }OT 

3.  100X;  or  100%  in  gold— 100  times  2.8X;  or  2.8%—  280/; 

or  280%  in  currency. 

/.  $1  in  gold  was  worth  $2.80  in  currency. 

(R.  3d  p.,  p.  217,prob.  8.) 


II. 

Ill, 
I. 


Bought  stock  at  10%  discount,  which  rose  to  5%  premium 
and  sold  for  cash.  Paying  a  debt  of  $33,  I  invested  the 
balance  in  stock  at  2%  premium,  which  at  par,  left  me 
$11  less  than  at  first;  how  much  money  had  I  at  first? 


86  FINKEL'S   SOLUTION  BOOK. 

(1.)         100%=my  money  at  first. 

(2.)         100%=par  value  of  stock. 

(3.)  10%=discount. 

(4.)         100%— 10%=90%=market  value. 

(5.)         .'.  90%=100%,  my  money;    because  that  is  the 

amount  invested. 

(6.)  l%=rv  of  100%=1£%,  and 

(7.)         100%=100  times  l£$,=lll£%=par  value  of  the 

stock  in  terms  of  my  money. 


(8.) 


II. 


/q  \ 


2.  1%—  l-|f%,and  [terms  ot  my  money. 

3.  5%—  5  times  11%—  5-f  %=premium  on  stock  in 

4.  llll%-|-5-t%=116f%  =  what  I  received  for  the 

stock. 

5.  116f  %  —  $33=amount  invested  in  second  stock. 

1.  100%=par  value  of  second  stock. 

2.  2%=premium. 

3.  100%-f-2%=102%=:market  valueof  second  stock. 

4.  116f%  —  $33=market  value  of  second  stock 

5.  .-.  102%=116J  %—  $33, 

6.  1%=^  of 

7.  100%=100  times 

$32T6T=par  value  of  second  stock. 

(10.)         114T\\%  —  $32T6T=what  1  received  for  the  second 

stock,  since  I  sold  them  at  par. 
(11.)         .'.  114TVij%—  $32T6T=100%—  $11,  by  the  last  con- 

dition of  the  problem. 
(12.)        1 


(13.)        i%__L^  of  $21T6T=$1.485,  and 
14T%% 

(14.)        100%=100  times  $1.485=$148.50. 
III.     .-.  I  had  $148.50  at  first.  (7?.  H.  A.,  p.  212,  prob.  8.) 

I.     Bought  $8000  in  gold  at  110%,  brokerage  •£%  ;    what  did 

I  pay  for  the  gold  in  currency? 
'1.   100%=par  value  of  gold. 

2.  110%=market  value. 

3.  -£%=brokerage. 

H.<[4.   110%+4%=110i%=  entire  cost. 

5.  100%— $8000, 

6.  1%=!-^  of  $8000=$80,  and 

7.  110l%=110^  times  $80=$8810=cost  of  gold  in  currency. 

III.     .'.  $8000  in  gold  costs  $8810  in  currency. 

I.  What  income  in  currency  would  a  man  receive  by  invest- 
ing $5220  in  U.  S.  5-20,  6%  bonds  at  116%,  when  gold 
is  worth  105? 


STOCKS  AND  BONDS.  87 

100%=par  value  of  the  bonds. 

116%— market  value. 
$5220— market  value. 
.-.  H6%  =$5220. 

1%— T|¥  of  $5220— $45. 

100%— 100  times  $45— $4500=par  value  of  bonds. 
100%— $4500. 

1%—^  of  $4500— $45. 

6%— 6  times  $45— $270— income  in  gold. 
$1.00  in  gold— $1.05  in  currency. 
$270  in  gold— 270  times  $1.05— $283.50  in  currency. 

$283.50— income  in  currency. 

(/?.  3d  p.,  p.  217,prob.  5.) 


What  %  of  income  do  U.  S.  4^-%  bonds,  at  108,  yield  when 
gold  is  105%? 

(1.)  100%— amount  invested  in  the  bonds. 

(2.)  100%— par  value  of  bonds. 

(3.)  108%=rnarket  value. 

(4.)  .-.  108%— 100%,  from  (1). 

(5.)  l%=riir  of  100%—  ff%,    [of  amount  invested. 

TT  I    (6.)  100%— 100  times  |-f-%=92$%— par  value  in  terms 
100%— 92-J%. 

4£%—  4^  times  ff%=4-^%— income  in  gold. 
8.)         100%  in  gold— 105%  in  currency. 
9.)  1%  in  gold— T-^¥  of  105%— l-fa%  in  currency. 

^(10.)  4-g-%  in  gold— ig-  times  l-2-1^%— 4f  %  in  currency. 

III.  .-.  Income  in  currency— 4f  %. 

Note. — This  is  a  general  solution  of  the  preceding  problem. 
Since  there  is  no  special  amount  given,  we  represent  the  amount 
invested  by  100%.  The  market  value  and  the  amount  invested 
being  the  same,  we  have  108%— 100%  as  shown  in  (4). 

I.     A  man  bought  Michigan  Central  at  120,  and  sold  at  124%;. 
what  %  of  the  investment  did  he  gain? 


II.< 


1.  124%—  selling  price. 

2.  120—  cost. 


3.  124%—  120%—  4%—  gain. 

4.  120%—  100%  of  itself. 

5.  l%=ThfOf  100%=f%, 

6.  4%—  4  times  |%=3£%=gain  on  the  investment. 


III.  .•.  He  gained  3£%  on  the  investment. 


I. 


II. 


FINKEL'S   SOLUTION  BOOK. 

What  sum  invested  in  U.  S.  5's  of  1881,  at  118,  yielded  an 
annual  income  of  $1921  in  currency,  when  gold  was 
at  113? 


$1.13  in  currency=$l  in  gold. 

$1  in  currency^^Vg.  of  $l=$i^|  in  gold,  and 

$1921  in    currency=1921   times 

come  in  gold. 
100%=par  value  of  the  bonds. 

5  %— income  in  gold. 
$1700=income  in  gold. 
.-.  6%= $1700, 

l%==\  of  $1700=$340,  and  [bonds. 

100%=iOO   times  $340=$34000=par  value  of  the 
100%=$34000, 

l%=^f  of  $34000=1340,  and 
118%=118  times  $340=$40120=market  value,  or 

amount  invested. 


III.     .'.  $40120=amount  invested. 


II. 


III. 


I. 


(I-) 
(2. 

(3. 

(4- 
(5. 
(6. 


(7.) 


(8.) 


(9.) 
(10.) 


(11.) 
(12.) 


SECOND    SOLUTION. 

100%=amount  invested  in  currency. 
100%=par  value. 
118%=market  value. 
...  118%=100%,  from  (1.) 

l%=rh-  of  100?fc=HHF%>  and 

100%=100    times    f|%=^84||%=par    value   in 
terms  of  the  investment. 


and 

d%=5  times  £f  %==4^%==income  in  gold. 
100%  in  gold— 113%  in  currency, 


in  gold=l-r'^        n  currency 
in  gold=4i|  times 
n  currency. 
$1921=income  in  currency. 


and 


of  $1921=$401.20,  and 

100%=100    times   $401.20=-$40120^amount    in- 
vested in  currency. 

$40120=amount  invested.        (JR.  3d  p.,  p.  218,  prob.  8.) 


How  many  shares  of  stock  bought  at  95^%,  and  sold  at 
105,  brokerage  \%  on  each  transaction,  will  yield  an 
income  of  $925  ? 


STOCKS  AND  BONDS. 


II. 


1.  100%=par  value  of  stock. 

2.  95J%=market  value  of  stock. 

3.  ^%==:brokerage. 

4.  95i+i%=95i%=entire  cost. 

5.  105%=selling  price-{-brokerage. 

6.  ^%=brokerage. 

7.  105%—  i%= 1041%=selling  price. 

8.  104|%—  < 

9.  $925=gain. 

10.  .-.  9i%: 

11.  1  %=~  of  $925=$100,  and 

ft* 

12.  100%=100  times  $100=$10000==pai  value  of  stock. 

13.  $100=par  value  one  share. 

14.  $10000=par  value  10000-f-lOO,  or  100  shares. 


III.  .-.  100=number  of  shares. 


(R.  3d  p.,  p.  218,prob.  9.) 


II. 


If  .1  invest  all  my  money  in  5%  furnace  stock  salable  at 
75%,  my  income  will  be  $180;  how  much  must  I  bor- 
row to  make  an  investment  in  5%  state  stock  selling  at 
102%,  to  have  that  income? 

1.  100%=par  value  of  furnace  stock. 

2.  5%=income. 

3.  $180=income. 

4.  .-.  5%=$180, 

5.  l%=i  of  $180=$36,  and  [nace  stock. 

6.  100% =100  times  $36=$3600=par  value  of  fur- 

1.  100%=$3600, 

2.  1%=T^  Ot  $3600=$36,  and  [nace  stock. 
$.     75%=75  times  $36=$2700=market  value  of  fur- 

1.  100%=par  value  of  state  stock. 

2.  6%=income. 


(!•) 


(8.) 


(5.) 


3.  $180=income. 

4.  .-.  6%=$180, 

5.  1%=^  of  $180=$30,  and  [stock. 

6.  100%=100  times  $30=$3000=par  value  of  state 

1.  100%  =$3000, 

2.  1%=T^  of  $3000=$30,  and  [state  stock. 

3.  102%=102    times    $30=$3060=market  value  of 
$3060— $2700=$360=what  I  must  borrow. 


III.  .-.  I  must  borrow  $360. 


(R.  H.  A.,  p.  225,  prob.  2. 


When  U.  S.  4%  bonds  are  quoted  at  106,  what  yearly  in- 
come will  be  received  in  gold  from  bonds  that  can  be 
bought  for  $4982  ? 


90  FINKEL'S   SOLUTION  BOOK. 

(1.)         100%— par  value  of  the  bonds. 
(2.)         106%— market  value. 

(3.)         $4982— market  value,  or  amount  invested. 
I  I    (4.)        .-.  106%— $4982, 

(5.)  1  %—  yfo  of  $4982— $47,  and 

(6.)         100%— 100  times  $47— $4700. 

rl.  100%— $4700, 
(7.K2-        1%—  Ti<5-  of  $4700— $47,  and 

13.       4%— 4  times  $47— $188=income  in  gold. 
JII.  .-.  $188— income  in  gold.  (fi.  3p.,  p.  218,  prob.  11.) 

I.  The  sale  of  my  farm  cost  me  $500,  but  I  gave  the  pro- 
ceeds to  a  broker,  allowing  him  ^%,  to  purchase  rail- 
road stock  then  in  the  market  at  102%;  the  farm  paid 
5%  income,  equal  to  $2075,  but  the  stock  will  pay 
$2025  more;  what  is  the  rate  of  dividend? 

(1.)         100%— value  of  the  farm. 

(2.)  5%— income  on  the  farm. 

(3.)         $2075— income  on  the  farm. 

(4.)        .-.  5%— $2075, 

(5.)  l%=i  of  $2075— $415,  and 

(6. )         100%— 100  times  $415— $41500— value  of  farm. 

(7.)  $41500— $500— $41000— amount  invested  in  stock. 
1.  100%— par  value  of  the  stock. 

2.  102%— market  value,  or  amount  invested. 

3.  ^%— brokerage 

4.  102%+^%— 102|%—  entire  cost  of  stock. 

fc=$41QOQ, 


of  $41000—  $400,  and 

[railroad  stock. 
7.  100%—  100  times  $400—  $40000—  par  value  of  the 

1.  $2075+$2025—  $4100—  income  on  railroad  stock. 

2.  $40000—100%, 

3.  $1%=^^  of  100%—^%  ,  and  [dend. 

4.  $4100—4100    times  =W=  rate    of  divi- 


III.  /.  10^%—  rate  of  dividend.  (/?.  H.  A.,  p.  224.,  prob.  4-) 

I.     What  must  be  paid    for  6%  bonds  to  realize  an  income  of 
8%  on  the  investment? 

1.  100%—  amount  invested. 

2.  6%—  income  on  the  par  value  of  the  bonds. 

3.  8  %—  income  on  the  investment. 

4.  .'.  8%  of  investment—  6%  of  the  par  value, 

5.  1%  of  investment—^  of  6%—  f  %  of  the  par  value,  and 
100%  of  investment—  100  times  f%—  75%  of  par  value. 

III.  .-.  Must  pay  75%  to  make  8%  on  the  investment. 

Note.  —  It  must  be  borne  in  mind  that  100%  of  any  quantity  is 
the  quantity  itself.    .-.  100%  of  the  amount  invested  equals  the 


II. 


STOCKS  AND  BONDS, 


91 


II. 


II. 


amount  invested.  It  must  also  be  remembered  that  the  income 
on  the  par  value  is  equal  to  the  income  on  the  investment.  Sup- 
pose I  buy  a  500-dollar  6%  bond  for  $400.  The  income  on  the 
par  value,  or  face  of  the  bond  is  6%  of  $500,  or  $30.  But  $30  is 
7%%  of  $400,  the  amount  invested.  Hence,  the  truth  of  step  4 
in  the  above  solution. 

Which    is    the  better    investment,    buying    9%    stock   at 
25%  advance,  or  6%   stock  at  25%  discount. 

1.)      100%—amount  invested  in  the  9%  stock, 
100%=par  value. 
25%==premium. 

100%+25%=125%= market  value. 
.-.  125%=100%, 

1%=^  of  100%=f  %,  and 
100%=100  times  f  %=80%=par  value  in  terms 
of  the  investment. 

1.  100%=80%, 

2.  I%=TO-H  of  80%=|%,  and  [stock. 

3.  9%=9   times   |%=:7i#==income    of    9% 
100%=amount  invested  in  6%  stock. 
100%= par  value  of  6%  stock. 

25%=discount. 

100%—  25%=75%=market  value. 
.-.  75%=100%. 

1%=TV  of  100%=li%,  and 

— 100   times    l^%=133^%=par  value  of 
the  6%  stock  in  terms  of  the  investment. 


LB. 


2.) 
(3. 
(4- 
(5. 
(6.) 
(7-) 


(8.) 


(2.) 
(3.) 
(4-) 
(5.) 
(6.) 
(7.) 


(8.)J2.       1%= 
IS.       6%= 


and  [stock. 

6%=6    times    l£%=8%=income    of  6% 
III.  .-.  The  latter   is   the  better  investment,  since  it  pays  8% — 
7.1.%  ?  or  \cjc  more  income  on  the  investment. 

(  Greenleafs  N.  A.,  p.  298,  prob.  5.) 

If  I  pay  87i%  for  railroad  bonds  that  yield  an  annual  in- 
come of  7%,  what  %  do  I  get  on  my  investment? 
100%=investment. 
100%=par  value. 

87^%=market  value,  or  amount  invested. 
.-.  87£%=100%,  from  (1.) 

1%— 1-  of  100%=1|%,  and 

o/'ff 

100%=100   times    l^%=114^%==par    value    in 
terms  of  the  investment. 

1.  100%=114f%, 

2.  1%==T^15.  of  114f  %=1^%,  and  [ment. 

3.  7%=7  times  l|%=8%=income  on  the  invest- 


(I-) 

(2.) 

It! 

(5.) 
(6.) 


(7.) 


Ill. 


)=income  on  the  investment. 


92 


FINKEL'S   SOLUTION   BOOK. 


A  banker  owns  2-J%  stocks  at  10%  below  par,  and  3% 
stocks  at  15%  below  par.  The  income  from  the  former 
is  66f  %  more  than  from  the  latter,  and  the  investment 
in  the  latter  is  $11400  less  than  in  the  former;  required 
the  whole  investment  and  income. 


II. 


!»J 

1. 

'2. 

(3.). 

4. 

5. 

6. 

rl. 

(4,)I: 

L 
1. 

2. 

3. 

(5.)< 

4. 
5. 

6. 

1. 

/A  J* 

(6-)   3. 

(8.) 

(9.) 

(10.) 

(11.) 

(12.) 

(13.) 

(14.) 


100%=investment  in  the  former. 

100% — $11400=investment  in  the  latter. 

100%=par  value  of  the  former. 
10%=discount  of  the  former,    [vested  in  former., 

100%—  10%=90%=market  value,  or  amount  in- 

.-.  90%=100%,  from  (1), 
1%=-^  of  100%=!-^%,  and 

100%=100  times  l£%=lll£%=par  value  of  for- 
mer in  terms  of  the  investment. 

100%=11H%, 

2-J-%=2£  times  11%— 2J%=income  of  former  in 

terms  of  the  investment. 
100%=par  value  of  the  latter. 

15%=discount.  [vested  in  the  latter. 

100% — 15%=85%=market  value,  or  amount  in- 
.-.  85%=100%— $11400,  from  (2), 

1%=^  of  (100%— $11400)=1T3T%— $134T2T, 
100%=100    times    (1T3T%—  $134T27)  =  117jj%— 

$13411^-3r=par  value  of  latter  in  terms  of  former. 

100%=117||%— $13411if,  [$134T2T,  and 

1%  =  T^  of  (H7fi.%— $l3411f3- )=  lT3y  %  — 
3%=3   times  (lfV% — $13412T)=319T% — $402T6T 
=income  of  latter  in  terms  of  the  investment. 
100%=income  of  the  latter. 
100%+66|%=166|%=income  of  the  former. 
2J%=income  of  the  former. 
/.  166f  %=2^%, 

of2|%=Jo%^nd 

[terms  of  income  of  former. 


"166* 


100%=100  times  ^%=lj%=income  of  latter  in 
3T9T%  —  $402T6T==income  oif  the  latter. 


, 
Iff  [former. 

100%=100    times    $216=$21600=investment    in 
100%—  $11400=  $21600  —  $11400  =$10200=-  in- 
vestment in  latter. 

=2^-  times  $216=$600=income  of  former. 


STOCKS  AND  BONDS. 


III. 


I. 


II. 


(15.)        aA-%— $402;fr=3JV   times  $216— $402-1<y=$36Q= 

income  of  latter. 

(16.)         $21600+$10200=$31800=whole  investment 
(17.)         $600+$360=$960= whole  income. 

J$31800=whole  investment,  and 
'•  |$960=whole  income.  (/?.  H.  A.,  p.  225,  prob.  4. ) 

W.  F.  Baird,  through  his  broker,  invested  a  certain  sum 
of  money  in  Philadelphia  6's  at  115^-%,  and  three  times 
as  much  in  Union  Pacific  7's  at  89-J%,  brokerage  -j-%  in 
both  cases;  how  much  was  invested  in  each  kincf  of  stock 
if  his  annual  income  is  $9920? 

(1.)         100%=amount  invested  in  Philadelphia  6's. 
(2.)         300%=amount  invested  in  Union  Pacific  7's. 


i. 
2. 

iuuy0=pai  vaiue  01  .rmiauejpma.  o  s>. 
115£%=market  value. 

3. 

-j-%—  brokerage. 

(3-). 

4. 

5. 

115-J%+|%=ll"6%=entire  cost  of  Phila.  6's. 
.r.  116%=100%. 

6. 

1%=^  of  100%=||%,  and 

7. 

100%—  100  timesff  %=  86A%=par  value  of  Phil- 

adelphia 6's  in  terms  of  investment. 

rl. 

100%  =86  2^%, 

MS 

1  %—  .j-i-g.  of  86^-%  ,  and 
6%=6    times    ff  %=5T5¥%=income  of   Philadel- 

I 

phia  6's  in  terms  of  investment. 

1. 

100%=par  value  of  Union  Pacific  7's. 

2. 

89^%=market  value. 

3. 

^.%—  brokerage. 

/  r    \ 

4. 

89|%-}-£^==90%==entire  cost  of  Union  Pacific  7's. 

(&r 

5. 

...  90%=300%, 

6. 

l%=fa  of  300%=3£%,  and 

7. 

100%=100    times    3|%=333i%=par    value    of 

Union  Pacific  7's. 

rl. 

100%=333£%, 

(6.)j2- 

-j  (ji  i     Q.C  QQQ  i  <T/.—  —  Q  JL  cL.    and 

1 

7%=7  times  3J%=23^%=income  of  Union  Pa- 

(7.) 
(8.) 
(9.) 

(10.) 

(11.) 
(12.) 


cific  7's  in  terms  of  investment. 

52^+23i%=28!r%=whole  income. 
$9920=whole  income. 


-  °f  ^9920 


[in  Philadelphia  7's. 
100%=100  times  $348=$34800=amount  invested 
300%=300   times   $348=1104600=5=    amount   in- 
vested in  Union  Pacific  7's. 


FINKEL'S   SOLUTION   BOOK. 


III. 


I. 


II. 


(  $34800=a mount  invested  in  Philadelphia  6's,  and 
I  $104400— amount  invested  in  Union  Pacific  7's. 

(R.  H.  A.,  p.  225,prob.  6.) 

Thomas  Reed  bought  6%  mining  stock  at  114^%,  and 
4%  furnace  stock  at  112% ,  brokerage  -£%  ;  the  latter 
cost  him  $430  more  than  the  former,  but  yielded  the 
same  income  ;  what  did  each  cost  him? 


(2.) 


(3.) 


(4.) 


100%=amount  invested  in  mining  stock. 
100%+$430=amount  invested  in  furnace  srock. 

1.  100%=par  value  of  mining  stock. 

2.  114^%=market  value. 

4.  ii4^%^-^%=1ilo%=ent\re  cost. 

5.  /.  115^=100%,  from  (1), 

6.  l%=Tl-g  of  100%— ff %,  and 

7.  100%=100     limes    |J%=965|^=par    vnlue    oi 

mining  stock  in  terms  of  investment. 


(5.) 


(6.) 

(7.) 
(8.) 


l%=rb  of  96ff  %=£$%,  and 
6%=6     times    |^%:==5-2\% 

stock  in  terms  of  investment. 
100%=par  value  of  furnace  stock. 
112%=market  value. 


of 


mining 


=~.  Tof  (100%+$430)=f%+$3it,  and 
-- 


100%=100    times    (|%-|_$3|l)_88|%-|-$382f= 
par  value  of  furnace  stock  in  terms  uf  invrestm't. 


2.  l%=rJT  of  (88f%+$382|  )=|%+|3|l,  and 

3.  4%=4   times  (|%+$3H)=  3|-  %+$15i|=income 

of  furnace  stock  in  terms  of  the  investment. 
.-.  5A%==s3f%.-fll5tti  by  the  conditions  of  the 
problem, 


!±a  of  $15= 


and 


(10.) 
(11.) 


[mining  stock. 
100%=100  times  $9,20=$920=amount  invested  in 
100%+|430=^1350=amount  invested  in  furnace 
stock.  (^?.  H.  A.,  p.  225,prob.  7.) 


Ill    *  /$920— amount;  invested  in  mining  stock,  and 
'  *|$1350=amount  invested  in  furnace  stock. 


STOCKS  AND  BONDS. 


95 


I. 


n. 


(i.) 


Suppose  10%  state  stock  is  20%  better  in  market  than  4% 
railroad  stock;  if  A.'s  income  be  $500  from  each,  how 
much  money  has  he  paid  for  each,  the  whole  investment 
bringing  6-gf  3-%  ? 

1.  100 %=par  value  of  state  stock. 

2.  10%=income. 

3.  $500=  in  come. 

4.  .-.  10% =$500, 

5.  1%=TL  of  $500=$50,and  [stock. 

6.  100%=100  times  $50=$5000=par  value  of  state 

1.  100%=par  value  of  railroad  stock. 

2.  4%=income. 

3.  $500=income. 

4.  .-.  4%=$500, 

5.  l%=i  of  $500=$125,  and  [railroad  stock. 

6.  100%=100    times    $125=$12500=par   value    of 
$5000=f  of  $12500,  i.  e.,  the   face   of  state  stock 

is  |-  of  face  of  railroad  stock. 

1.  100%=whole  investment. 

2.  G^l-g  %=in  come  of  whole  investment. 

3.  $500+$500=$1000=income  of  whole  investment. 


(2.) 


(3.) 


(4.) 


(5.) 


6. 


1%=— L-  of  $1000=$166.50,  and 

6^  [ment. 

100%=100  times  $166.50=$16650=whole  invest- 


1.   100%=investment  in  railroad  stock. 

l'.  40%=|  of  100%=investment  in  state  stock, 

excluding  the  20%  excess. 
2'.  100%  =40%, 
3'.  l%=T^o  of  40%=|%,  and 
4.  20%=20    times    |%  =8%=excess   of    state 

stock  over  same  amount  of  railroad  stock. 

3.  40%+8%=48%=investment  in  state  stock. 

4.  100%+48%=148%=whole  investment. 

5.  $16650=whole  investment. 

6.  .-.  148%=$16650, 

7.  l%=T^g-  of  $16650=$112.50,          [railroad  stock. 

8.  100%=100  times  $112.  50=$11250=investment  in 

9.  48%=48    times    $112.50=$5400=investment    in 

state  stock 

j.        ^  J$11250=amount  invested  in  railroad  stock,  and 
'  '  '  1  $5400=amount  invested  in  state  stock. 

(R.  H,  A,,  p.  227,  prob.  5.) 


9d  FINKEL'S   SOLUTION  BOOK. 


EXAMPLES. 

1.  What  could  I  afford  to  pay  for  bonds  yielding   an   annual 
income  of  9%  to  invest  my  money  so  as  to  realize  6%  on  the  in- 
vestment? Ans.  150%. 

2.  What  must  I  pay  for  Chicago,  Burlington  &  Quincy  Rail- 
road stock  that  bears  6%  that  my  annual  income   on   the   invest- 
ment may  yield  5%  ?  Ans.  120%. 

3.  Bought  75  shares  N.  Y.,  P.  &  O.  Railroad  stock  at  105%, 
and  sold  them  at  108£%  ;  how  much  did  I  gain  in  the  transac- 
tion? Ans.  $262.50. 

4.  How  many  shares   of  bank   stock   at  5%    premium,  can  be 
bought  for  $7665?  Ans.  73. 

5.  A  broker  bought  stock   at  4%   discount,  and,  selling  them 
at  3%  premium,  gained  $1400  ;  how  many  shares  did  he  buy? 

Ans.  200. 

6.  At  what  price  must  I  buy  15%  stock  that  it  may  yield  the 
same  income  as  4%  stock  purchased  at  90%  ?  Ans.  337-^%. 

7.  How   much   must  I  pay  for  New  York   6's  so  that  I  may 
realize  an  income  of  9%?  Ans.  66f  %. 

8.  At  what  price  must  I  buy  7%  stock  so  that  they  may  yield 
an  income  equivalent  to  10%  stocks  at  par?  Ans.  70%. 

9.  What  sum  must  I  invest  in  U.  S.  6's  at  118%  to  secure  an 
annual  income  of  $1800 ?r  Ans.  $35400. 

10.  Which  is   the   more  profitable,  and  how  much,  to   invest 
$5000  in  6%  stock  purchased  at  75%,  or  5%  stock  purchased  at 
60%?  Ans,    The  latter;  $16|. 

11.  If  a  man  who  had  $5000  U.  S.  6's  of  1881  should  sell  them 
at  115%,  and  invest  in  U.  S.  10-40' s  purchased  at  105%,  would 
he  gain  or  lose  and  how  much?  Ans.  Loss  $26.19. 

12.  When  gold  is  at  120,  what  is  a  "greenback"  dollar  worth  ? 

Ans. 


13.  Suppose  the  market  value  of  5%  bank  stock  to  be  'll-g-% 
higher  than  8%  corporation  bonds  ;  I  realize  8%   on  my  invest- 
ment, and  my  income  from  each  is  $180;  what  did   I   invest   in 
each?  Ans.  $2923.07 A  in  former,  and  $1576.92T\  in  latter. 

14.  A  bought  5%  railroad  stock  at  109|%,  and  4|%  pike  stock 
at  107-J%,  brokerage  |% ;  the  former  cost  $100  less  than  the  latter 
but  yielded  the  same  income;  what  did  each  cost  him?     „ 

Ans.  $1100  cost  of  former,  and  $1200  cos»  of  latter 


INSURANCE.  97 

15.     What  rate  %  of  income  shall  I  receive  if  I  buy  U.  S.  5's  at 
a  premium  of  10%,  and  receive  payment  at  par  in  15  years? 

Ans.     " 


16.  Suppose  the  market  value  of  6%  corporation  stock  is  20% 
less  than  5%  state  stock;  if  my  income  be  $1200  from  each,  what 
did  I  pay  for  each  if  the  whole  investment  brings  6%  ? 

Ans.  $16000,  and  $24000. 


17.  I  bought  2|%  stock  at  80%,  and  4|%  stock  at 
The  income  on  the  former  was  44f  %  more  than  on  the  latter, 
but  my  investment  is  $22140  less  in  the  latter  than  in  the  former; 
what  do  I  realize  on  my  investment?  Ans. 


Hint. — Find  the  whole  investment,  and  whole  income  as  in  the  problem 
on  page  75.  Then  find  what  %  the  whole  income  is  of  the  whole  invest- 
ment. 

18.  Invested   in   U.  S.  4£'s  at   105,  brokerage   £%  ;  f  as  much 
in  U.  P.  6's  at  119|,  brokerage  £%;  and  3  times  as  much  in  N. 
Y.  7's,  at  87J,  brokerage  \%.     If  my  entire  income  is  $1702,  find 
my  investment.    (School  Visitor,  vol.  12,  p.  97.)          Ans.  $25320. 

19.  A.  paid  $1075  for  U.  S.  5-20  6%  bonds  at  7|%  premium, 
interest  payable  semi-annually  in  gold.     When  the  average  pre- 
mium on  gold  was  112%,  did   he  make  more  or  less  than  B.  who 
invested  an  equal  sum  in  railroad  stock  at  14%  below  par,  which 
paid  a  semi-annual  dividend  of  4%? 

Ans.    A.  makes  $16.40  less  than  B.  every  six  months. 

20.  I  invested  $4200  in  railroad  stock  at  105,  and  sold  it  at 
80%;  how  much  must  I  borrow  at  4%  so  that  by  investing  all  I 
have  in  6%   bonds   at  8%    interest,  payable  annually,  I  may  re- 
trieve my  loss  in  one  year?  Ans.  $18600. 


VII.     INSURANCE. 

1.     Insurance  is  indemnity  against  loss  or  damage. 

~  T  M.  Fire  Insurance. 

1.  Property  Insurance,  j  2    j.^  Insurance 


2.     Insurance. 


2.  Personal  Insurance. 


1.  Life  Insurance. 

2.  Accident  Insurance. 


3.  Health  Insurance. 

3.  Property  Insurance  is  the  indemnity  against  loss  or 
damage  of  property. 

4.  Personal  Insurance  is  indemnity  against  loss  of  life 
or  health. 

5.  Fire  Insurance  is  indemnity  against  loss  by  fire. 


98  FINKEL'S  SOLUTION  BOOK. 

6.  Marine  Insurance  is  indemnity  against  the  dangers 
of  navigation 

7.  Life  Insurance  is  a   contract    in    which   a   company 
agrees,  in  consideration   of  certain    premiums   received,  to  pay  a 
certain  sum  to  the  heirs  or  assigns  of  the  insured  at  his  death,  or 
to  himself  if  he  attains  a  certain  age. 

8.  Accident  Insurance  is    indemnity    against    loss    by 
accident. 

9.  Health  Insurance  is  a  weekly  indemnity  in   case  of 
sickness. 

1C.    TJie   Insurer,  or    Underwriter,  is  the  party,  or 
company,  that  undertakes  the  risk. 


11.  The  Itisfc  is   the   particular   danger  against  which  the 
insurer  undertakes. 

12.  The  Insured  is  the  party  protected  against  loss. 


13.  The  frennium  is  the  sum  paid   for  insurance;  and  is 
a  certain  per  cent,  of  the  amount  insured. 

14.  The  Amount,  or   Valuation,  is  the  sum  for  which 
the  premium  is  paid. 


I.  My  house  is  permanently  insured  for  $1800,  by  a  deposit 
of  ten  annual  premiums,  the  rate  per  year  being  |%; 
how  much  did  I  deposit,  and  if,  on  terminating  the  in- 
surance, I  receive  my  deposit  less  5%  ;  how  much  do  I 
get? 

(1.)         100%=$1800, 

(2.)         I%=TT><T  of  $1800=$18,  and 

II. 


\&. )          j.  /o:==^-^  wi  tp±ouu=<j>io,  anu 

(3.)         i%=|  times  $18=$13.50=one  annual  deposit. 

(4.)         $135=10  times  $13.50=ten  annual  deposits. 

/•I.  100%=$135, 
(5.){2.  1%=^  of  $135=$1.35,  and 

13.  5%=5  times  $1.35=$6.75=deduction. 
(6.)        $135— $6.75=$128.25=what  I  received. 

ry    .    (  $135=amount  deposited,  and 
'  '  '  (  $128.25=amount  received. 

(7?.  H.  A.,  p.  230,  prod.  5.} 

I.  An  insurance  company  having  a  risk  of  $25000,  at  T9^%, 
reinsured  $10000,  at  \% ,  with  another  office,  and  $5000, 
at  1%,  with  another;  how  much  did  it  clear  above  what 
it  paid  ? 


INSURANCE. 


99 


II. 


(1.)  100%—  $25000, 

(2.)  l%=T-b  of  $25000—  $250,  and 

(3.)  T9(r%=T9<r  times   $250—  $225—  what   the   company 
received  for  taking  the  risk. 

1.  $10000—  amount  the  company  reinsured  at  f%. 

2.  100%—  $10000, 

(4.)    3.  1%=  -rfoof  $10000—  $100,  and 

4.  f%—  f  times  $100—  $80—  what  the  company  paid 
for  reinsuring  $10000. 

1.  $5000—  amount  reinsured  in  another  office  at  1%. 

2.  100%—  $5000,  [for  reinsuring  $5000. 
(5.)    3.  1%—  T|-g-  of  $5000—  $50—  what  the  company  paid 

4.  $80+$50—  $130—  what  the  company  paid  out. 

5.  $225—  $130—  $95—  what  it  cleared. 

III.  .-.  $95—  what  the  company  cleared. 

(/?.  H.  A.,  p.  230,prob.  7.) 

I  took  a  risk  at  4-J%  ;  reinsured  -|  of  it  at  2%,  and  £  of  it 
at  2-^%  ;  what  rate  of  insurance  do  I  get  on  what  is  left? 


(1.) 

(2.) 


100%—  whole  risk. 

—  premium. 

—  |  of  100%^amount  reinsured  at  2%. 


?— y^-g-  of  40%— 1%,  and      [suringf  of  therisk. 
%— 2  times  -|  %— 1^%— ampunt  I  pay  out  for  rein- 
fl.  25%—  i  of  100%— second  part  reinsured. 
2.  100%— 25%. 
(4.K8.  1%— T^r  of  25%— i%,  and 

4.  2^-%— 2-J-  times  ^%—|%— amount  I  paid    out   for 
(          reinsuring  £  of  the  risk. 

(5.)         |%-|-f%—l^J%— amount  of  premiums  paid  out. 
(6.)         11%—1^%—^%—amountof  premium  I  had  left. 
(7.)         40%-f-25%— 65%— whole  amount  reinsured. 
(8.)         100%— 65%—  35%—  risk  left  on  which  I  received 

TQ%  premium. 
fl.  35%— 100%  of  itself. 
(9.K2,  1%—  ^  of  100%—  2f%,  and 

(.3.  ^ofa—.^  times  2^%=T3-?.%=r=rate  of  premium 

III.     ,'.  T3?%— rate  of  insurance  I  receive. 

(R.  H.  A., p.  281,prob.  6.) 

Remark. — 35%  is  the  base  and  -fa%  is  the  percentage,  and  we 
wish  to  know  what  per  cent,  -}$%  is  of  35%. 

I.  Took  a  risk  at  2%;  reinsured  $10000  of  it  at  2^%  and 
$8000  at  1|%;  my  share  of  the  premium  was  $207-50; 
what  sum  was  insured? 


100  FINKEL'S  SOLUTION  BOOK. 

rl.  100%=$10000, 

(1.K2.  1%=  -r^of  $10000=4100,  and   [$10000 reinsured. 
13. 


r%=2^  times  $100=$212.50=amount  paid  out  on 

rl.  100%=$8000, 

(2.K2.  l%=Ti¥of$8000=$80,and         [$8000  reinsured. 

13.  If  %=lf   times  $80=$140=amount   paid  out  on 

(3.)  $212.50+$140=$352.50=whole  amount  paid  out. 

(4.)  $207. 50=what  I  realize. 

(5.)  .'.  $352.50+207.50=$560=premium  on  whole  risk. 

(6.)  100%=risk. 

(7.)  2%=premium. 

(8.)  $560=premium. 

(9.)  .'.  2%=$560, 

(10.)  l%=i  of  $560=$280,  and 

(11.)  100^=100  times  $280=$28000=risk. 

III.     .-.  $28000=risk.  (R.  H.  A.,  p.  232,  prob.  6.) 


I.  I  can  insure  my  house  for  $2500  at  T8¥%  premium  annually, 
or  permanently  by  paying  down  12  annual  premiums; 
which  should  I  prefer,  and  how  much  will  I  gain  by  it 
if  money  is  worth  6%  per  annum  to  me? 

r  (1.)         100%=$2500. 
(2.)        l%=Tta  of  $2500=$25,  and 
(3.)        -nr%=T8¥  times  $25=$20=one  annual  premium. 
(4.)         $240=12  times  $20=twelve  annual  premiums. 

1.  100%=the    amount   that   will    produce   $20   an- 

nually at  6%. 

2.  6%=iriterest. 
!!.<!    ft.  J3.  $20=interest. 

l°'^4.  /.  6%=$20, 

5.  l%=iof$20=$3i,  and 

6.  100%=100   times  $3i=$333^=the  amount  that 

will  produce  $20  annually  at  6%. 

(6.)         $3331H-$20=$3531i=amount  j  would  have  to  pay 
down  by  the  former  condition.  [tion. 

L  (7.)         .',  $353i~$240=$113i=gain  by  the  latter  condi- 

TTT        (  The  latter  is  the  better. 
'  I  $113i=gain. 

Remark. — In  (6)  we  add  $20,  since  a  payment  must  be  made 
immediately.  $333£  will  not  produce  that  sum  until  the  end  of 
the  year. 


INSURANCE. 


The  Mutual  Fire  Insurance  Company  insured  a  building 
and  its  stock  for  f  of  its  value,  charging  If  %.  The 
Union  Insurance  Company  relieved  them  of  £  of  the 
risk,  at  l-J-%.  The  building  and  stock  being  destroyed 
by  fire,  the  Union  lost  $49000  less  than  the  Mutual; 
what  amount  of  money  did  the  owners  of  the  building 
and  stock  lose? 

100%=value  of  the  building  and  stock. 
66f%=f  of  100%=amount  insured. 
lj%=rate  of  insurance. 
100%=66f%, 

nnr  of  66f  %=f 
=1|  times  f%= 

from  the  owners  of  the  building  and  stock. 
16f  %=i  of  66|%=amount  of  which  the  Union 
relieved  the  Mutual. 


=         and 


r  (i-) 

SI 


and 
=what  Mutual  received 


(10.) 

(11.) 

(12.) 
(13.) 

(14.) 
(15.) 

(16.) 

(17.) 
(18.) 


(21.) 


of 
1-J%=1-J.  times  -|-%— ^%— what  the  Mutual  paid 

the  Union  for  taking  the  risk  of  16|%. 
16f  %+li%=17-f%=whole  amount  the  Mutual 

received.  [paid  out. 

amount   the   Mutual 
i^,~=  amount    the   Mutual 

lost."  ' 

16f  %=amount  the  Union  paid  the  Mutual. 
^%=amount  the  Union  received  from  the  Mutual. 
.'.  16f  % — i%=16T5T%=amount  the  Union  lost. 
491V%— 16T\%=  32f%=what    the    Mutual   lost 

more  than  the  Union.  [Union. 

$49000=what    the    Mutual     lost    more    than    the 
...  32f  %=$49000, 

1  %= JL  of  $49000=$1500,  and 
'       32f  [ing  and  stock. 

100%=100  times  $1500=$  150000= value  of  build- 

66|%=66|-  times  $1 500=$  100000=a mount  in- 
sured, [ers  lost,  it  not  being  insured. 

33j-%=33£  times  $1500=$50000=what  the  own- 

100%  =$100000, 

1%=-^  of  $100000=$1000,  and 

l|%=l|  times  $1000=$1750=what  the  owners 
paid  the  Mutual  for  insurance. 

.-.  $50000 +  $1750=  $51750=whole  amount  the 
owners  lost. 


III.  .'.  The  owners  of  the  building  and  stock  lost  $51750. 


102  FINKEL'S   SOLUTION  BOOK. 


EXAMPLES. 

1.  At  l-f%,  the   premium    for   insuring   my  stare  was  $89.10; 
what  was  the  amount  of  the  insurance?  Ans.  $6480. 

2.  The  premium  for  insuring  a  tannery  for  f  of  its  value,  at 
,  was  $145.60;  what  was  the  value  of  the  tannery? 

Ans.  $11648. 

3.  A  store  and  its  goods  are  worth  $6370.     What  sum  must 
be  insured,  at  2%,  to  cover  both  property  and  premium? 

Ans.  - 

4.  The  premium   for   insuring  $9870  was  $690.90  ;  what  was 
the  rate?  Ans.  7%. 

5.  A  merchant  whose  stock  of  goods  was  valued  at  $30000, 
insured  it  for  f  ot  its  value,  at  f  %.     In  a  fire  he  saved  $5000  of 
the  goods.     What  was  his  loss?     What  was   the   loss  of  the  in- 
surance companies?  Ans.          • 

6.  A  man  paid  $180   for  insuring   his   saw  mill   for  f-  of  its 
value  at  3%;  what  was  the  value  of  the  mill?  Ans.  —  . 

7.  A  house  which  has  been  insured  for  $3500  for  10  years,  at 
\°/c  &  year,  was  destroyed  by  fire  ;  how  much  did   the  money  re- 
ceived from  the  company  exceed  the  cost  of  premiums? 

Ans.  —  —  . 

8.  Took  a  risk  on  a   house  worth  $40000,  at  2%;  reinsured 
•J  of  it  for  2^%,  and  ^  of  it  at  2-J%;  in  each  case  the  amount  cov- 
ers premium;  how  much  do  I  gain?  Ans.  $99.558* 

9.  Took  a  risk  at  If  %;  reinsured  f  of  it  at  2J%  ;  my  share 
of  the  premium  was  $43  ;  what  was  the  amount  of  the  risk? 

Ans.  $17200. 


10.  Took  a  risk  at  2£%  ;  reinsured  -J  of  ?.t  at  a  rate  equal  to 
3%  of  the  whole,  by  which  I  lost  $37.50.  What  was  the  value 
of  the  risk?  Ans.  $5000. 


SIMPLE  INTEREST.  103 

CHAPTER   XIII. 

INTEREST. 

I.     SIMPLE  INTEREST. 

1.  Interest  is  money  paid  by  the  borrower  to  the   lender 
for  the  use  of  money. 

2.  The  Principal  is  the  sum  of  money  for  which  interest 
is  paid. 

3.  The  Rate  of  interest  is  the  rate  per  cent,  on  $1  for  a 
certain  time. 

4.  The  Time  is  the  period  during  which  the  money  is  on 
interest. 

5.  The  Amount  is  the  sum  of  the  principal  and   interest. 

6.  Simple  Interest  is  interest  on  the  principal  only. 

7.  Legal  Interest  is  at  the  rate  fixed  by  law. 

Usury  is  interest  at  a  rate  greater  than  that  allowed  by- 


8. 
law. 


Let  P=the  principal, 

r=the  interest  on  $1  for  one  year, 
7?=l-[-r=a mount  of  $1  for  one  year, 
7Z=the  number  of  years, 
yl=amount  of  P  for  n  years, 
/V=simple  interest  on  P  for  a  year, 
Pnr— simple  interest  on  P  for  n  years. 
P -\-Pnr— P  (l-|-?zr)=amountof  P  forn  years, 
of  P  for  n  years. 

(I.); 


~\+nr*       *v      ' 
...  Pnr=A—P. 

A — P     Interest 


When  any  three  of  the  quantities  A,  P,  n,  r  are  given,  the 
fourth  may  be  found. 

CASE  I. 
rPrincipal,^ 

Given<  Rate,  and  >to  find  the  interest.     Formula,  7=s/V«. 
iTime,        J 


104  FINKEL'S  SOLUTION  BOOK'. 

I.     Find  the  interest  of  $300  for  two  years  at  6%. 
By  formula, 

Interest  /V«=$300X-06x2=$36. 
By  100%  method. 

rl.  100%=$300, 

1 2.  l%=Tfc>  of  $300=$3,  and. 

13.  6%=6  times  $3=$18=interest  for  one  year. 
U.  $36=2  times  $18=interest  for  2  years. 

III.     .'.  $36=interest  on  $300  at  6%  for  2  years. 

CASE    II. 

(Principal,  >>  A—/> 

Given<Rate,  and  >to  find  the  time.     Formula,  n= — = — . 

llnterest,    )  Pr 

I.     In  what  time,  at  5%,  will  $60  amount  to  $72? 
By  formula, 

A—P    $72—  $60 


By  100%  method. 

1.  $72=amount. 

2.  $60=principal. 

3.  $72— $60=$12=interest  for  a  certain  time. 
II.<J4.  100%  =$60, 

5.  i%==_|^  of  $60=$f,  and 

6.  5%=5  times  $f=$3=interest  for  one  year. 

7.  $12=interest  for  12-4-3,  or  4  years. 

III.  V.  $60  at  5%  will  amount  to  $72  in  4  years. 

CASE  III. 

rPrincipal,^  A—P 

Given^  Time,  and  >to  find  the  rate.     Formula,  r=— - — . 

llnterest,    ) 
I.     I  borrowed  $600  for  two  years  and  paid  $48  interest;  what 

rate  did  I  pay? 
By  formula, 

A—P      I 


By  100%  method. 
'1.  $48=interest  for  2  years. 
2.  $24=£  of  $48=interest  for  1  year. 
II.J3.  $600=100%, 

4.  $l=:^oflOO%=i 
[5.  $24=24  times  i-%=4 
III.     .'.  I  paid  4%  interest. 


SIMPLE  INTEREST:  105. 


rTime,         ^ 

Given<[Rate,  and  >to  find  the  principal. 
llnterest.     J  Fo 


CASE  IV. 
rTime, 

A  —  P      1 
Formula,  P== 

I.     The  interest  for  3  years,  at  9%,  is  $21.60;  what  is  the 
principal  ? 

By  formula, 

P_A-P_  7_$21.60     . 
~-~-- 


II. 


By  100%  method. 

1.  $21.60=interest  for  3  years. 

2.  $7.20=4  of  $21.60=interest  for  1  year. 

3.  100%=principal. 

4.  9%=interest  for  1  year. 


5.  $7.20=interest  for  1  year. 

6.  .-.  9%=$7.20, 

7.  l%=i  of  $7.20=$.80,  and 
100%=100  times  $.80==$80=principal. 

III.     .'.  $80=the  principal. 


CASE   V. 
Given<(Rate,  and>to  find  the  principal.     Formula,  P= 


fTime,        i  A 

LAmount    J  l+nr' 


I.     What  principal  will  amount  to  $936  in  5  years,  at  6%  ? 

By  formula, 

A  $936 

-=  -  06~ 


By  100%  method. 

1.  100%=principal. 

2.  6%=interest  for  1  year. 

3.  3Q%=5  times  6%=interest  for  5  years. 

4.  100%+30%=130%=amount. 

5.  $936=amount. 

6.  .-.  130%=$936, 

7.  1%=^  of  $936=$7.20,  and 

8.  100%=100  times  $7.20=$720=principal. 

III.     .'.  $720=the  principal  that  will  amount  to  $936  in  5  years 
at  6%. 


106  FJNKEL'S  SOLUTION  BOOK. 

I.     In  what  time  will  any  sum  quadruple  itself  at 

1.  100%=principal.     Then 

2.  400%=the  amount 

3.  .-.  400%—  100%=300%=interest. 

4.  8%=interest  for  1  year. 

5.  300%=interest  for  300-=-8,  or  37£  years. 


III.     .-.  Any  principal  will  quadruple  itself  in  37-J  years  at 


II.     TRUE  DISCOUNT. 

1.  Discount  on   a   debt   payable  by  agreement  at  some 
future  time,  is  a  deduction  made  for  "cash,"  or  present  payment; 
and  arises  from  the  consideration  of  the  present  'worth  of  the  debt. 

2.  Present  Worth  is   that  sum  of  money  which,  put  oa 
interest  for  the  given  time  and  rate,  will  amount   to  the  debt  at 
its  maturity. 

3.  True  Discount  is  the  difference  between  the  present 
worth  and  the  whole  debt. 

Since  P  will  amount  to  A   inn  years,  P  may  be  considered 
equivalent  to  A  due  at  the  end  of  n  years. 

.*.  P  may  be  regarded  as  the  present  worth  of  a  given  future 
sum  A. 

•  P—  *- 

~~ 


I.     Find  the  present  worth  of  $590,  due  in  3  years,  the  rate 

of  interest  being  6%. 

By  formula, 

A  $590 

=-a= 


By  100%  method. 

1.  100%=present  worth. 

2.  6%=mterest  on  present  worth  for  1  year. 

3.  18%=3x6%=interest  for  3  years. 

TT  ;4.  100%+18%=118%=amount,  or  debt. 
<5.  $590=debt. 

6.  .-.  118%=$590, 

7.  1%—yfg-  of  $590=$5,  and 

8.  100%  =100  times  $5=$500=present  worth. 

HI. .'.  $500=present  worth  of  $590  due  in  3  years  at 


BANK  DISCOUNT. 


107 


I.  A  merchant  buys  a  bill  of  goods  amounting  to  $2480;  he 
can  have  4  months  credit,  or  5%  off  for  cash  :  if  money 
is  worth  only  10%  to  him,  what  will  he  gain  by  paying 
cash? 


•••i ;,! 

) 


(2. 
(3. 
(4. 

(5. 
(6. 

(7. 

(8. 
(9. 


(11.) 


III.  .'.  He 


100%=present  worth  of  the  debt. 
10%=interest  on  present  worth  for  1  year. 
3-j.%=interest  for  4  months. 
100%+3i%=103i%=amount  of  present  worth, 

which  equals  the  debt,  by  definition. 
$2480=the  debt. 
...  103i%=$2480, 

1  %=?7^T  of  $2480=$24,  and 
106$ 

100% =100  times  $24=$2400=present  worth. 

$2480— $2400=$80=true  discount. 

100%=$2480. 

1  %=Tfo  of  $2480=$24.80,  [count  for  cash. 

5%=5  times  $24.80=$124=trade  discount,  or  dis- 

.'.  $124 — $80=$44=his  gain  by  paying  cash. 

would  gain  $44  by  paying  cash. 

(/?.  3d  p.,  p.  258,prob.  10.) 


1. 


Remark.— It  is  clear  that  $2480— $124,=$2356  would  pay  for 
the  goods  cash.  If  the  merchant  had  this  sum  of  money  on  hand, 
it  would,  in  4  months,  at  10%,  produce  $78.53-j-  interest.  But  if 
he  pays  his  debt  he  will  make  $124.  Hence  he  will  gain  $124 — 
$78.534=$45.46f. 


III.     BANK  DISCOUNT. 


1.  J$anh  Discount  is  simple   interest  on   the   face  of  a 
note,  calculated  from  the  day  of  discount  to  the  day  of  maturity, 
and  paid  in  advance. 

2.  The  Proceeds  of  a  note  is  the  amount  which  remains 
after  deducting  the  discount  from  the  face. 

CASE  I. 

(  Face  of  note,  ) 
Given  <  Rate,  and        >  to  find  the  discount  and  proceeds. 

f  Time.  S  T^         i       {  Z?=^X^X» 

v  '  Formulae,  j  P==:j7_jr)t 


108  FINKEL'S  SOLUTION  BOOK. 

I.     What  is  the  bank  discount  of  $770  for  90  days,  at  6%  ? 
By  formula, 

=$770  X  -06  X       ±^=$1  1  .935. 


By  100%  method. 

II.  100%=$770, 
2.  l%=T-^of$770=$7.70,  and 
3.  6%=6  times  $7.70=$46.20=discount  for  1  year. 
4.  $11.935=^  of  $46.20=discount  for  93  days. 

III.  /.  $11.935=bank  discount  on  $770  for  90  days  at  6 


CASE  II. 
ds,    ) 
and  >  to  fi 

(    Rate» 


C  Proceeds,    ) 
Given  <  Time,  and  >  to  find  the  face  of  the  note. 


TT  77 

Formula,  F=-  -  . 
1—  rn 

I.     For  what  sum   must  a   note  be   made,  so   that  when  dis- 
counted at  a  bank,  for  90  days,  at  6%    the  proceeds  will 
be  $393.80? 
Bv  formula, 

P  $393.80 


By  100%  method. 

1.  100%=face  of  the  note. 

2.  6%=discount  for  one  year. 

3.  1^%=^  of  6%=discount  for  93  days. 

4.  100%—  lH%=98<nr%=proceeds. 

5.  $393.80=proceeds. 

7.  1%=——  of  $393.80=$4,  and 


II. 


5.  100% =100  times  $4=$400=face  of  the  note. 
III.  /.  $400=face  of  the  note. 

CASE  III. 

Given  rate  of  bank  discount,  to  find  the  corresponding  rate  of 

interest.  Formula,  rate  of /.== . 

1 — rn 

I.     What  is  the  rate  of  interest  when  a  60  day   note  is  dis- 
counted at  8%  per  annum? 
By  formula, 


of  7== 


ANNUAL  INTEREST.  109 


II. 


By  100%  method. 
4.   100%=face  of  note. 

2.  8%=discount  for  1  year. 

3.  12^=  e^.  of  8%=discount  for  63  days. 


4.  100%— If  %=98f  %—proceeds. 


5.  98|%=100%  of  itself. 

6.  1%=—-  of  100%— Jff %,  and. 

~f,  o%=8  times  £$£%=81yg-%=rate  of  interest 
III.     .-.  The   rate  of  interest  on  a  60  day   note   discounted  at  8% 
per  annum=8^%%. 

CASE   IV. 

Given    the    rate  of   interest,  to  find  the  corresponding  rate  of 

,,  .  rate  of  /. 

discount  Formula, 


I.     What  is  the  rate  of  discount  on  a  60  day  note  which  yields 
10%  interest? 

By  formula, 


By  100%  method. 
.  100%=proceeds. 

2.  10%=interest  on  proceeds  for  1  year. 

3.  11%=^  Of  10%=interest  on  proceeds  for  63  days. 

4.  100%+l|%=101|%=face  of  note. 

5.  101|%=100%  of  itself. 


II. 


7.  10%=1 
III.  "  .-.  The  rate  of  discount— 

Note.  —  It  must  be  borne  in  mind  that  the  interest  on  the  pro 
ceeds  is  equal  to  the  discount  on  the  face  of  the  note. 


IV.    ANNUAL    INTEREST. 

1.  A.in/nAJWJ&  Interest  is  the  simple  interest  of  the  princi- 
pal and  each  year's  interest  from  the  time  of  its  accruing  until 
settlement. 

I.  No  interest  having  been  paid,  find  the  amount  due  Sept. 
7,  1877,  on  a  note  of  $500,  dated  June  1,  1875,  with 
interest  at  6%,  payable  semi -annually. 


no 


FINKEL'S  SOLUTION  BOOK. 


2  yr.  3  mon. 


1  yr.  9  mon.  6  d. 


1  yr.  3  mon.  6  d. 


9  mon.  6  d. 


1st 


2d 


3m.6d 


3d 


4th 


6d 


II. 


(6.) 


(1.)  100%=$500,  1877—9—7 

(2.)  1%=T^  of  $500=$5,  and  1875—6 — 1 

(3.)  6%=6 times $5=$30=simple interest         2—3 — 6 

for  1  year. 
(4.)   $68=2T\    times    $30=simple    interest    for   2  years,  3 

months,  6  days. 
(5.)   $15=J  of  $30=semi-annual  interest. 

1.  100%=$15,  ' 

2.  l%=Tfoof$15=$.15,  and 

3.  6%=6    times    $.15=$.90=interest    on    one    semi- 

annual interest  for  1  year. 

4.  $3.885=4^-f  times  $.90=interest  on  one  semi-annual 

interest  for  the  sum  of  the  periods  each  draws  int. 
(7.)  V.  $500+$68+$3.885=$571.885=amount  of  the  note. 
III.  V.  $571.885=amount  of  the  note. 

Explanation. — At  the  end  of  six  months  there  is  $15  interest 
due  ;  and,  since  it  was  not  paid  at  that  time,  it  drew  interest 
from  that  time  to  the  time  of  settlement,  which  is  1  yr.  9  mon. 
6  da.  At  the  end  of  the  next  six  months,  or  at  the  end  of  the 
first  year,  there  is  another  $15  due;  and,  since  it  was  not  paid  at 
that  time,  it  drew  interest  from  that  time  to  the  time  of  settle- 
ment, which  is  1  yr.  3  mon.  6  da.  In  like  manner,  the  third 
semi-annual  interest  drew  interest  for  9  mon.  6  da.,  and  the 
fourth  for  3  mon.  6  da.  This  is  the  same  as  one  semi-annual 
interest  drawing  interest  for  the  sum  of  1  yr.  9  mon.  6  da., 
1  yr.  3  mon.  6  da.,  9  mon.  6  da.,  3  mon.  6  da.  In  the  dia- 
gram, the  line  A  B  represents  2  yr.  3  mon.  6  day.,  A  1  repre- 
sents the  first  year  the  note  run,  and  1-2  represents  the  second 
year  the  note  run.  Between  A  and  1  is  a  small  mark  that  de- 
notes the  semi-annual  period  ;  also  one  between  1  and  2.  By 
such  diagrams,  the  time  for  which  to  compute  interest  on  the 
simple  interest  may  be  easily  found. 

I.  The  interest  of  U.  S.  4%  bonds  is  payable  quarterly  in 
gold;  granting  that  the  income  from  them  might  be 
immediately  invested,  at  6%,  what  would  the  income 
on  20  1000-dollar  bonds  amount  to  in  5  years,  with  gold 
at  105? 


ANNUAL  INTEREST. 


Ill 


Syr. 


$200  for  4  yr.  9  mon.  at  6^, 


$200  for  4  yr.  6  mon.  at  656. 


$200  for  4  yr.  3  mon.  at  6$. 


2d     3d  I  4th 


&c. 


10     11      12 


16      17      1 


II. 


(i- 

(2. 


{I: 

(v. 


$1000=par  value  of  one  bond. 

$20000=par  value  of  20  bonds. 

100%=$20000, 

1%=T^  of  $20000=$200,  and 

4%=4  times  $200=$800=income  for  one  year. 

$4000=5  times  $800=income  for  five  years. 

$200=i  of  $800=interest  due  at  the  end  of  first 
quarter,  and  which  draws  interest  to  time  of  set- 
tlement. 

100%=$200, 

1%=-^  of  $200=$2,  and  [est  for  one  year. 

6%=6  times  $2=$12=interest  on  quarterly  inter- 

$570=47-^  times  $12=interest  on  quarterly  inter- 
est for  the  sum  of  4f  yr.-f-4-^-  yr.-f-4^  yr.-j- 

-|- i  yr. ,  or  47-J  years. 

/.  $4000+$570=$4570=income  of  bonds  in  gold. 

$1.00  in  gold=$1.05  in  currency.  [rency. 

$4570  in  gold=4570  times  $1.05=$4798.50  in  cur- 

III.  .-.  The  bonds  yield  $4798.50  in  currency. 


'1. 

2. 

(8.). 

3. 
4. 

(9.) 

(10.) 

1(11.) 

Explanation. — It  must  be  borne  in  mind  that  the  quarterly  in- 
terest, $200,  is  put  on  interest  at  6%  as  soo'n  as  it  is  due.  At  the 
end  of  the  first  quarter  there  is  $200  due  which  draws  interest  at 
6%  for  the  remaining  time,  4  years,  9  months.  The  second 
quarterly  interest  is  due  at  the  end  of  six  months  and  draws  in- 
terest for  the  remaining  time,  4  years  3  months,  and  so  on  with 
the  remaining  quarterly  payments.  This  is  the  same  as  one 
quarterly  payment  drawing  interest  for  the  sum  of  4|  yr.-|-4J  yr. 
yr.+etc.,  or  47-^-  years. 


I.     What  was  due  on  a  note  of  $1200,  dated  January  16,  1883, 
and  due  Aug.   1,    1892,     and    bearing    interest    at    8, 
payable  annually,  if  the  2,  3,  5,  and  7th    years' 
were  paid  ? 


interest 


112 


FINKEL'S   SOLUTION   BOOK. 


$96  for  9  yr.  6  m.  15  d.  at  85*. 


$96  for  6  yr.  6  m.  15  d. 


4yr.  6m.  15  d. 


$96,  2yr.  6m.  15  d. 


1  yr.6  m.  15  d 


15  d 


II., 


(8.) 


(9.) 
1(10) 


100%=$1200. 

1%=^  of  $1200=$12,  and 

8%=8  times  $12=$96=simple  int.  for  one  year 
$480=5  X$96=five  simple  interests. 
=J  of  $96=interest  for  6  months. 

$48=interest  for  15  days. 
.*.  $532=simple  interest  unpaid. 

1.  100%=$96. 

2.  l%=Ti?  of  $96=$.96,  and  [simple  interest. 

3.  8%=8  times  $.96=$7.6S=interest    on    one  year's 

4.  $193.92=25i  times  $7.68=interest  on   year's  sim- 

ple interest  for  9  yr.  6  mon.  15  da.,+6  yr.  6  mon. 
15  da.,+4  yr.  6  mon.  15  da.,-f-2yr.  6  mon.  15  da., 
+1  yr.  6  mon.  15  da., -(-6  mon.  15  da.,  or  25  yr. 
3  mon. 

.'.   $532+$193.92=$725.92=amount  of   interest 
due.  [1,  1892. 

$1200+725.92  =  $1925.92  =  amount     due     July 


III.  .-.  $1925.92=whole  amount  due  Aug.  1,1892. 
V.     COMPOUND  INTEREST. 

1.      Compound  Interest  is  interest  on  a  principal  formed 
by  adding  interest  to  a  former  principal. 

Let  />=principal  on  compound  interest. 


./?:=(  l-j-/')=amount  of  one  dollar  for  1  year. 
P  (l+r)=/)7?=amount  of  P  dollars  for  1  year. 
P  (l-|-r)2=/>^?2=amount  of  P  dollars  for  2  years. 
P  (l+/)3==^?3=amount  of  P  dollars  for  3  years. 
P  (l-[-r)n=f>l?n=amount  of  P  dollars  for  n  years. 
Let  ^4=amount  of  P  dollars  in  n  years,  and 

/—the  compound  interest  of  P  dollars  for  »  years. 
Then  I==P^—P  .....  I. 
.  II. 


.III. 


COMPOUND  INTEREST.  113 

n 

.*.  JR=\/T IV.    Applying  logarithms  to 

p 

n  log.  /?=log.  A — log.  P ',  whence 
_log.  A— log.  P 

log./? 

When  compound  interest  is  payable  semi-annually. 
P  (1-f-f  )=amount  of  P  dollars  for  4-  year. 
p  (!-(-!-)  2=amount  of  P  dollars  for"l  year. 
P  (l-j-f)2n=amount  of  P  dollars  for  n  years. 
.-.  A=P  (l-f-J)2n,  when  payable  semi-annually. 
When  compound  interest  is  payable  quarterly, 
P  (1-f-f  )=amount  of  P  dollars  for  £  year. 
P  (1-j-f  )2=amount  of  P  dollars  for  -J-  year. 
P  (1-j-f  )3=amount  of  P  dollars  for  £  year. 
P  (1-j-f  )4=amount  of  P  dollars  for  1  year. 
P  (1-j-f  )4n— amount  of  P  dollars  for  n  years. 


When  the  interest  is  payable  monthly, 
A-£(1+A)»««. 

When  the  interest  is  payable  q  times  a  year, 
A=P  (l+|)qn- 

CASE  I. 

(  Principal,  } 
Given  <  Rate,  and  >  to  find  the  compound  interest  and  amount. 

(  Time,         )  (  I=PR*—P, 

Formula,, 


I.     Find    the  compound  interest  and  amount  of   $500   for  3 

years  at  6%. 
By  formulas, 

^=/>7?n=:$500X(l+.06)3=$595.508,  and 
I=PR*—  />=$500  X  ( 1+.06) 3—  $500=$95.508. 
Remark. — In    compound  interest,  the   100%  method  becomes 
very  tedious. 

By  100%  method. 
'  (1.)        100%=$500, 

(3.)         6%=6  times  $5= $30=interest  for  1  year. 
(4.)         $500+$30-=$530=arnount,  or   principal    for  the 
second  year. 

..  100%=$530, 

2.  1  %=r^r  of  $530=45.30,  [year. 


rIL^ 


(5.) 


3.  6%=6    times   $5.30=$31.80=interest  for  second 

4.  $530+$31.80=$561.80=amount,  or  principal  for 

the  third  year. 


114 


FINKEL'S  SOLUTION  BOOK. 


III. 


1.  100%=$561.80, 

2.  1  %=Tfor  of  $561.80=45.618,  and  [year. 
(6.)    3.  6%=6   times  $5.618— $33.70S=interest  for  third 

4.  561.80+$33.7.08=$595.508=amount  at  end  of  the 

third  year. 
(7.)X      $595.508— $500=$95.508=compound  interest. 

(  $95.508=compound  interest,  and 
'"'  (  $595.508=compound  amount. 


CASE  II. 


C  Principal, 
Given  <  Rate,  and 

(  Compound  Interest, 


to  find  the  time. 
Formula,  n= 


log.  A— log.  P 


log./? 

I.     In  what  time  will  $8000  amount   to   $12000,   at  6%  com- 
pound interest? 
By  formula, 

log.  A— log.  JP=log.  12000— log.  8000^ 
log.  R  log.  1.06 

4.079181—3.903090     _  1K  .... 

=6  yr.    11   mon.    15   da.      We  may   solve  the 

problem  thus:  $8000(  1.06  )n=$  12000,  whence  (1.06)n=12000-j- 
8000=1.50.  Referring  to  a  table  of  compound  amounts  and 
passing  down  the  column  of  6%,  we  find  this  amount  between  6 
years  and  7  years. 

The  amount  for  6  years  is  1.4185191 ;  the  amount  for  required 
time  is  1.50.  .'.  There  is  a  difference  of  1.50— 1.4185191,  or 
.0814809.  The  difference  for  the  year  between  6  and  7  is 
.0851112.  .0851112— amount  for  the  whole  period  between  6 
and  7,  .0814809—amount  for  g%tiif  of  the  period  or,  11  mon.  15 
da.  /.  The  whole  time=6  yr.  11  mon.  15  da. 


CASE  III. 

(  Principal,  ^ 

Given  <  Compound  Intesest  or  Amount,  and  >  to  find  the  rate. 
(  Time,  ) 

Formula,  r=n  l^ 1. 

I.     At  what  rate,  by  compound   interest,   will   $1000  amount 
to  $1593.85  in  8  years? 
By  formula, 


ANNUITIES.  115 


CASE  IV. 

C  Compound  Interest  or  Amount  } 

Given  ]  Time,  and  V  to  find  the  principal 

(  Rate,  \ 


I.     What    principal,    at    compound    interest    will  amount  to 
27062.85  in  7  years  at  4%  ? 
By  formula, 


CHAPTER   XIV. 

ANNUITIES. 

I.  An,  Annuity  is  a  sum  of  money  payable  at  yearly,  or 
other  regular  intervals. 

!1.  Perpetual,  or 
I:  SSKor 
4.  Contingent. 

3.  A  Perpetual  Annuity  is  one  that  continues  forever. 

4.  A  Limited  Annuity  ceases  at  a  certain  time. 

5.  A    Certain  Annuity  begins  and  ends  at   fixed  times. 

6.  A  Contingent  Annuity  begins  or  ends   with   the 
happening  of  a  contingent  event. 

7.  An  Immediate  Annuity  is  one  that  begins  at  once. 

8.  A  Deferred  Annuity  is  one  that  does  not  begin  im- 
mediately. 

9.  The  Final  or  Forborne  value  of  an  annuity  is  the 
amount  of  the  whole  accumulated  debt  and  interest,   at  the  time 
the  annuity  ceases. 

10.  The    Present     Value    of   an    annuity    is    that    sum, 
which,  put  at  interest  for  the   given   time   and   given   rate,  will 
amount  to  the  initial  value. 

II.  The  Initial    Vallie  of  an  annuity  is   the   value  of  a 

deferred  annuity  at  the  time  it  commences. 


116 


FINKEL'S  SOLUTION  BOOK. 


to  find  the  initial  value  of  a  perpetuity. 


CASE  I. 
C  Annuity,     ) 
Given  <  Time,  and 

(  Rate, 

I.  What  is  the  initial  value  of  a  perpetual  annuity  of  $300  a 
year,  allowing  interest  at  6%  ? 

1.  iOO%=initial  value. 

2.  6%=interest  for  1  year. 

3.  $300=interest  for  1  year. 

4.  .-.  6%=$300. 

5.  \°/c=\  of  $300=$50,  and 

§.  100%=100  times  $50=$5000=initial  value. 
III.  ^  .-.  Initial  value=$5000.  (R*  H.  A.,  p.  310,  prob.  1.) 

I.  What  is  the  initial  value  of  a  perpetual  leasehold  of  $2500 
a  year  payable  quarterly,  interest  payable  semi-annually 
at  6%;  6%  payable  annually  ;  6%  payable  quarterly? 

1.  Let  5=the  annuity.     Then  6'=the  amount  due  in 

3  months. 

2.  6'-)-5'(l-|-^)=namount  due  in  6  months. 

3.  /.    v4  =  5+6'(l-f-.01i-)=:$625  +  $625(1.01|)  = 

$1259.37i=amount  due  at  the  end  of  6  months. 

4.  100%— initial  value. 

5.  3%=semi-annual  annuity. 

6.  $1259.37^=semi-annual  annuity. 


HJ 


B.< 


8.  l%=Jof  $1259.37i=$419.7916f  ,  and 

9.  100%=100  times  $419.7916f=initial  value. 

1.  Let  .Sr^amoiint  due  in  3  months.     Then 

2.  •S-|-5l(l-f-j)==ainount  due  in  6  months,  [and 

3.  614-61(l4-T)+^(l+|r)=amount  due  in  9  months, 

4.  S+S(l+-J)+S(l+}r)+S(l+Jr  )==  amount   due 
in  1  year.  [(1+'V)^2556.25. 

5.  .-.  ^=$625+$625(  l+-\6)+  $625(1+'  V2)+  $625- 

6.  100%=5nitial  value. 

7.  6%=annuitv. 

8.  $2556.25=annuity. 

9.  .-.  Q%=  $2556.25. 

10.  1%=4  of  $2556.25=$426.0416|,  and  [value. 

11.  100%=100  times  $426.0416f  =$42604.  16f=initial 

1.  100%==initial  value. 

2.  l-^%=quarterly  annuity. 

3.  $625=quarterly  annuity. 


5.  1^=      of  $625-4416.6666|,  and 

6.  100%=100  times  $416.6666|=$41666. 


ANNUITIES. 

Initial  value  of  A=$41979.16f, 
Initial  value  of  B=:$42604.16f , 
Initial  value  of  C=$41666.66f . 

(R. 


117 


CASE  II. 


Given 


Annuity, 
Interval, 
Rate,  and 


and 


H.  A.,  p.S10,prob.5.) 


to  find    the   present 
value  of  a  deferr- 


Time  the  perpetuity  is  deferred,  J       ed  perpetuity. 


Let  6*=the  annuity,  f— the  rate,  and  R=l-\-r.  Then  by  Case 
I.,  the  initial  value  of  S  is  S-~-r.  To  find  the  present  value  of 
the  initial  value,  we  use  formula  III.,  compound  interest.  .-.  P 

S  S 

au    ,.  , — rr=  nu  D    -^-r-  in   which  /  is  the  time   the  perpetuity 
r  (\-\-rJ-     R'(R — 1) 

is  deferred. 

I.     Find  the  present  value  of  a  perpetuity  of  $250  a  year,  de- 
ferred 8  years,  allowing  6%  interest. 
By  formula, 

$250  _      $250 

~R*(R—  1)     (l-f-.06)8(l+.06— 1)~~.06(1.06)8~~ 
By  100%  method. 
(1.)   100%=initial  value. 
2.)  6%=annuity. 
3.)  $250=annuity. 
4.)  /.  6%=$250. 
5.)  l%=i  of  $250=$41|,  and 
!!.<!  (6.)  100%=100  times  $41f=$4166.66£=mitial  value. 

1.  100%=present  value  of  $4166.66|  due  in  8  years 

at  6%. 

2.  159.38481  %==(  1.06)  8XlOO%=compound  amount 

of  the  present  value  for  8  yr.  at  6%. 

3.  /.  159.38481  %=$4166.66f, 

4.  l^^i-^Wr  of  $4166.66f=$26. 1422,  and 


III. 


I. 


5.  100%=100   times   $26.1422  =  $2614.22  =  present 

value. 

/.  The  present  value  of  a  perpetuity  of  $250  a  year  He- 
ferred  8  years  at  6%  interest=$2614.22. 

Find  the  present  value  of  an  estate  which,  in  5  years,  is 
to  pay  $325  a  year  forever;  interest  8%,  payable  semi- 
annually. 

By  formula, 

5  $325  $325 


"[ 
$2 


.0816(1.04)10 


690.67. 


118  FINKEL'S   SOLUTION  BOOK. 

By  100%  method. 
(1.)   100%— initial  value. 
(2.)  4%— amount  due  in  6  months. 
(3.)  4%+(1.04)x4%— 8.16%— amount  due  in  1  year. 
(4.)   $325— amount  due  in  1  year. 
(5.)  .-.  8.16%— $325, 

(6.)   1%=^  of  $325— $39.828431,  and  [value. 

(7.)   100%— 100    times    $39.828431=  $3982.8431  =  initial 

1.  100%— present  value  of  $3982.8431. 

2.  148.024428f*:=(1.04)10X100%==compound  amount 

of  100%  for  5  yr.  at  8%. 


II. 


1(8.) 


3.  .-.148.024428%— $3982.8431, 


4.  l%=T48.<ji44?8°f  $3982.8431—  $26.9067,  and 

5.  100%—  100  times    $26.9067  —  $2690.67  —  present 

value. 
III.     .'.  $2690.67—  present  value  of  the   estate. 

(JR.   H.   A.,  p.  311,prob.  4.) 

Explanation.  —  The  initial  value  is  a  sum  of  money  which 
placed  on  interest  at  8%  payable  semi-annually  will  produce 
$325  per  year.  But  8%  payable  semi-annually  is  the  same  as 
8.16%  payable  annually.  Hence  8.16%  is  the  annual  payment. 
But  $325  is  the  annual  payment.  Hence  8.16%—  $325,  from 
which  we  find  that  $3982.8431  is  the  initial  value,  or  the  amount 
that  will  produce  $325  per  year.  Then  the  present  value  of  a 
sum  of  money  that  will  pay  $325  is  $3982.8431  if  the  payments 
are  to  begin  at  once,  but  $3982.8431-r-  (1.04)10  if  the  payments 
are  not  to  begin  until  the  end  of  5  years. 

CASE  III. 


Given  -<  £.nnul,ty>  ,  Uo  find  the  present  valve  of  an  an- 

I  lime  to  run,  and  I  .,  ~.    . 

T    .         T  nuity  certain. 

(^Interval,  ) 

(a)  Let  P  denote  the  present  value.     The  amount  of  P  for  n 


Let  6*—  the  payment,  or  amount  due  the  first  year. 
=  the  amount  due  the  second  year. 

-  the  amount  due  the  third  year. 

S'-|-S1/?--»S7?2+S7?8==the    amount   due    the    fourth 
year.  [due  the  nth  year. 

S+  Sfi+Sfi'*+  Sfi*+  .......  +  67?"-1  —  amount 

.-.  A= 


...(1) 
AR=  SR 


(2),  by  multiplying  (1)  by  R. 
AR—A=SR*—S.  .  .  (3),  by  subtracting  (1)  from  (2). 

1)  •  •  •  •    4.       But  PR*=A. 


ANNUITIES.  119 


•'"  (5')>   whence 

- 

- 


When  the  annuity  is  to  begin  at  a  certain  time,   and  then 
to  continue  a  certain  time. 

Let/— the  number  of  years  the  annuity   is   deferred,  and  q= 
the  number  of  years  the  annuity  continues.     Then 

,        S         ~Rp+q 1 

P=-= — -  X    T?p+q — =the  present  value  of  an  annuity  S,  for 

the  time  (p-\-y)  years,  and 

C  Dp 1 

P"=— — -X — D^— =the  present  value   of  an  annuity   S,  for  p 


R  P+V  )      R — 1      R  P+V 

I.     Find  the  present  value  of  an  annuity  of  $250,  payable  an- 
nually for  30  years  at  5%. 
Given  S,  n,  and  r. 
By  formula, 

S         R«-\     $250     (1.05)so-l     ,0040110* 
^7?=rX~7F~  r05~X    (1.05)30    : 
By  100%  method. 
(1.)         100%— initial  value. 
(2.)         5%— annuity. 
(3.)         $250— annuity. 
(4.)        .-.  5%—  $250, 
(5.)        l%-=i  of  $250—$50,  and 
(6.)        100%— 100  times  $50%=$5000=initial  value  of 

an  immediate  perpetuity  of  $250  per  year. 
'1.  100% —present  value  of  an  annuity  deferred  30 
years.  [ent  value  for  30  years. 

2.  432.19424%— (1.05)30XlOO%— amount   of  pres- 

3.  .-.  432.19424%— $5000, 

4.  l%=¥T7^Trre  of  $5000— $11.568865,  and 

5.  100%=100  times  $11.568865— $1156.8865— pres- 

ent value  of  annuity  of  $250  deferred  30  years. 
(8-)  "      /.  $5000— $1156.8865=$3843.1135— present  value 

of  an  annuity  continuing  30  years. 
.!II.  V.  $3843.1135— present  value  of  an  annuity  of  $250,  payable 
annually  for  30  years, 


i20 


FINKEL'S   SOLUTION  BOOK. 


Remark. — Since  $5000  is  the  initial  value  which,  in  this  case, 
is  also  the  present  value  of  an  immediate  perpetual  annuity,  or 
perpetuity  of  $250,  and  $1156.8865  the  present  value  of  an  an- 
nuity of  $250  deferred  30  years,  $5000— $1156.8865— $3843.1135= 
the  present  value  of  an  annuity  of  $250  continuing  for  30  years 


I. 


II. 


Find  the  present  value  of  an  annuity  of  $826.50,  to  com- 
mence in  3  years  and  run  13  years,  9  months,  interest 
6%,  payable  semi-annually. 

Given  £=$826.50,  r=.06,  /=3  years,  and  ^=13|  years. 
When  interest  is  payable  semi-annually,  _/?=(l-[-|-)2. 
By  formula  (7), 

$826.50..  a0609)^1=$6324m 


'       .0609   '       (1.0609)16^ 


(8.) 


(9.) 


(10.) 


III. 


By  100%  method. 
(1.)         100%=initial  value. 
(2.)         3%=amount  due  in  6  months. 
(3.)         3%+3%  (1.03)=6.09%=amount  due  in  1  year. 
(4.)         $826.5Q=amount  due  in  1  year. 
(5.)        .-.  6.09%=$826.50, 
(6.)         l%=Tfo  of  $826.50=$135.712643,  and 
(7.)         100%  =  100   times   $135.712643  =  $13571.2643= 
initial  value 

1.  100%=present  value   of  a   perpetuity  of  $826.50 

deferred  3  years. 

2.  119.40523  %=(  1.0609  )2   times   100%=amount   of 

present  value  for  3  years. 

3.  .-.  11940523%=$13571.2643, 

4.  l%=mr.ArB7Tr°f  $13571.2643=$113.6686, 

5.  100%=100  times  $113.6586=$11365.86=present 

value  of  such  a  perpetuity  deferred  3  years 

1.  100%=present  value  of  such  a  perpetuity   deferr- 

ed 16f  years. 

2.  269.212027%=(1.0609)16^  times   100%  =  amount 

of  present  value  for  16f  years 

3.  /.  269.212027%=$13571.2643,  . 

4.  1%=^^^  of  $13571.2643=150.4117, 

5.  100%=100  times    $50.4117=  $5041.17=  present 

value  of  such  a  perpetuity  deferred  16f  years. 

.-.  $11365.86— $5041. 17=$6324.69=present  value 
of  an  annuity  of  $826.50  deferred  3  years  and 
continuing  13|  years. 

$6324.69=present  value  of  $826.50,  etc. 


If  the 


ANNUITIES. 


121 


If  the  annuity  is  to  begin  in  p  years  and   continue   forever,  the 
formula, 

S 


if    ^=00,    the 


For,  since    P= 
quantity 


=1 =1 — 0,  approaches  1  as  its  limit, 

CO 


and  we  have  ^ 


I.     Find  the  present  value  of  a  perpetual  annuity  of  $1000  to 
begin  in  3  years,  at  4%  interest. 

By  formula,  [value  of  the  annuity. 

8  $1000 


II, 


By  100%  method. 
(1.)         1 00  %=initial  value. 
(2.)         4%=annuity. 
.)         $1000=annuity. 


3 

(4-) 

(5.)         l%=i  of  $1000=$250,  and  [$1000. 

(6.)         100%=100  times  $250=$25000=initial   value  of 

1.  100%=present  value. 

2.  112.4864%  =  (1.04) 3    times    100%  =  amount   of 

present  value  for  3  years  at  4%. 

3.  .-.  112.4864  %=$25000, 

4.  l%=nnriimrof  $25000=$222.2492,  and 

5.  100%=100  times  $222.2492=$22224.92=present 

value. 

III.  .-.  $22224.92=present  value  of  an  annuity  of  $1000  to  be- 
gin in  3  years  at  4%. 


f  Annuity, 
J  Rate, 
\  Interval,  and 


CASE  IV. 
-to  find  the  final  or  forborne  value. 


Given 

' 

(^Tirne  to  run, 
Let  $=amount  due  first  year. 

$_|_l$fjff=amount  due  second  year. 

nt  due  third  year. 

=:amount  due  the  fourth  year. 

2+  &ff3  + +o'7?n-1=  amount  due 

the  nth  year. 
Let  ^4=amount  due  the  ^th  year. 

-1  ...  (1). 


122  FINKEL'S   SOLUTION  BOOK. 


.  .  (2),  by  multiplying  (1)  by  R.  [from  (2). 

AR—  A=Sfi"—  S  ......  (3),  by  subtracting   (1) 


A  pays  $25  a  year  for  tobacco  ;  how  much  better  off  would 
he  have  been  in  40  years  if  he  had  invested  it  at  10% 
per  annum? 

By  formula, 


—  1]  =  $11064.8139. 


II. 


By  100%  method. 

1.  100%=initial  value. 

2.  10%=annuity. 

3.  $25=annuity. 

4.  .-.  10%=$25, 

5.  1%=TV  of$25=$2.50,  and 


6.  100%=109  times  $2.50=$250=initial  value. 

7.  $44.2592556=[(1.10)40—1]X$1— compound  interest  of 

$1  for  40  yr.  at  10%.  [$250  for  40  yr.  at  10%. 

I  8.  .'.  $11064.8139=44.2592556  X$250=compound    int.   of 
III.  /.  He  would  be  $11064.8139  better  off. 

Remark. — $250  placed  on  interest  at  10%  will  produce  $25 
per  year.  If  this  interest  be  put  on  interest  at  10%,  instead  of 
spending  it  for  tobacco,  it  will  amount  to  $11064.8139  in  40 
years.  This  would  be  a  very  sensible  and  profitable  investment 
for  every  young  man  to  make,  who  is  a  slave  to  the  pernicious 
habit. 

I.     An  annuity,  at  simple  interest  6%,  in  14  years,  amounted 
to  $116.76  ;  what  would  have  been  the  difference,  had  it 
been  at  compound  interest  6%  ? 
(1.)         100%=initial  value,  or  the   principal  that   would 

produce  the  annuity. 
(2.)         6%=annuity  for  1  year. 
(3.)         84%=14x6%=annuity  for  14  years. 

1.  100%=6%, 

2.  1%=T^  of  6%=-^Q-%,  and  [1  year. 
(4.)    3.  6%=6  times -^r%=2T%=interest  on  annuity  for 

4.  32.76 %=91   times   2T%=interest  on   annuity  for 

(1+2+3+ +14),  or  91  years. 

(5.)  84%+32.76%=116.76%=whole  amount  of    the 
n.<  annuity. 

(6.)  $116.76=whole  amount  of  the  annuity. 

(7.)  /.  116.76%=$116.76, 

(8.)  1%=^^  of  $116,76=$!,  and 

(9.)  100%=100  times  $l=$100=initial  value. 

(10.)  6%=6  times  $l=$6=annuity. 


ANNUITIES. 


122 


(11.)         $1.260904—  [(1.06) I*— l]x$l=  compound  inter- 
est on  $1  for  14  yrs.  at  6%. 

(12.)         $126.0904— 1.260904 X $100 —compound    interest 
on  $100  for  14  yrs.  tit  6%. 

(13.)         .-.  $126.0904— $116.76— $9.3304— difference. 
III.  .-.  The  difference=|9.8304. 


CASE  V. 
Final  Value  or  Present  Value 


Given  <  Rate,  and 

(  Time  to  run, 

c< 


Solving  ^— 


r>n  _  -J 


to  find  the  annuity. 


R— 1 


with  respect  to  6*  and  we  have 
T (1).     If  ^l=the  final  or 


forborne  value,  by  the    formula   in   the   last   case,  we   have   A- 
~™ — 1.      Solving  this  with  respect  to  6",  we  have. 


S=- 


(2). 


I.     How  much    a  year  should   I   pay,  to   secure  $15000  at  the 

end  of  17  years,  interest  7%  ? 
By  formula  (2), 

rA         .07  X  $15000 
o—  Ou -=— — —  =  $4ob.o& 

By  100%  method. 

(1.)         100%— annuity. 

(2.)         7%— annuity. 

(3.)         .-.  7%— 100%, 

(4.)         l%=y  of  100%— 14f%,  and 

(5.)         100%— 100  times  14f%— 1428f  %— initial  value. 

1.  100%— present  value  of  1428^%  due  in  17  years. 

2.  315.8815%— amount  of  present  value  for  17  years. 

4.  i%=-jfTT.foTT  of  142T8|%=4.522591%,  and 

5.  100%— 100  times  4.522591%— 452.2591%— pres- 
II. <  ent  value. 

(7.)         .'.  1428^%  —452.2591%  —  976.3223%  —  present 

value  of  an  annuity  running  17  years. 
(8.)         3.1588152%— (1.07)17  times  l%=amount  of  1% 

for  17  years. 
(9.)        3084.0217%— (1.07) 17  times  976.3223 %=amount 

of  976.3223%  for  17  years  at  7%. 
(10.  $15000— amount,  or  final  value. 

(11.          .-•  3084.0217%— $15000. 
(12.          l%=inrreWTT  of  $15000— $4.8638,  and 
*(  13.          100%=100  X  $4.8638— $486.38— annuity. 


124  FINKEL'S   SOLUTION  BOOK. 

III.  .-.  I  must  pay  $486.38. 

CASE  VI. 

(  Annuity,  )  ,  ~  ,  ,.  .. 

s^.  i  T»  i  TT  i  c  J.T-  A  •*.  j  i  ^o  nna  time  it 

Given  <  Present  Value  of  the  Annuity,  and  } 

)  T-»  *  \  runs. 

(  Rate,  ) 

In   formula  (6),  Case  III.,  we  have  P=— — -x — ^ — ,  whence 

Pr_S—Pr 

S~   ~~S~* 

.•.^?n=— — ^— (1).     Applying  logarithms, 


S      \  log.  5—  log.  (S—Pr) 

--    -  ~ 


I.     In   how   many  years   can   a   debt  of  $1,000,000,  drawing 
interest  at  6%,   be   discharged    by   a   sinking   fund   of 
$80,  000  per  year? 
By  formula  (2), 
^Aog.  s—  log.  (  S—  Pr  )==\og.  80000—  log.  (  80000—1000000  X  .06  ) 

log.  R  log.  1.06 

log.  80000—  log.  20000^4.903090—  4.301030_.6Q2060:= 
log.  1.06  .025306  ~  025306" 

years. 

By  another  method. 

Assume  $1,000,000  to  be  the  present  value  of  an  annuity  of 
$80000  a  year.  Then  $12.50  may  be  considered  as  the  present 
value  of  $1  for  the  'same  time  and  rate.  By  reference  to  a  table 
of  present  worths  $12.50,  which  is  1000000-f-80000,  will  be 
found  to  be  between  23  and  24  years. 

Note.  —  A  table  of  present  worths  may  be  computed  by  form- 
ula (6.),  Case  III.,  in  which  put  6'=$1. 

I.     In  what  time  will  a  debt  of  $10000,    drawing   interest  at 
6%,  be  paid  by  installments  of  $1000  a  year. 
By  formula, 

log.  S—  log.  (S—  7V)_log.  1000—  log.  (  1000—10000  X.  06) 
log.  R  log.  1.06 

3—2.602060     1KI70-  1K 

-=15.725  years=15  yr.  8  mo.  21  da. 


.025306 

By  another  method. 

Assume  $10000  to  be  the  present  value  of  an  annuity  of  $1000 
a  year.  Then  $10000-r-1000=$10=the  present  value  of  $1  for 
the  same  time  and  rate.  By  referring  to  a  table  of  present  worth 
we  find  this  amount  between  15  and  16  years.  .*.  The  time  is  15 
years  -f- 


ANNUITIES.  125 

The  compound  amount  of  $10000  for  15  yr.  at  Q%=    $23965.58 
The  final  value  of  $1000  for  15  years  at  6%=  $23275.97 

Balance^    $     689.61 

This  balance,  $689.61,  will  require  a  fraction  of  a  year  to  dis- 
charge it.  The  part  of  a  year  required,  will  be  such  a  fraction  of 
a  year  as  the  amount  of  $689.61  for  \\\e  fraction  of  a  year  is  of 
$1000. 

6%  of  $689.61  for  fae  fraction  of  a  year=$41.3766X  fraction 
of  a  year. 

.-.    $689.61+$41.3766X  fraction   of  a    year=the   amount   of 
$689.61    for   the  fraction   of  a   year.     This   amount   divided  by 
$1000,  a  yearly  payment,  will  give  infraction. 
$689.61+141.3766  ^fraction 

-$1655~  =/'**>*  whence 

$689.61+141.3766  X  fraction=$\(yM  X  fraction 

.«.  $1000  yJraction—$±\.  3766  X/™c#o«==$689.61,  or 


689.61      0 

=&  months,  19  days. 


.•.  The  whole  time=15  yr.  8  mon.  19  da. 

CASE  VII. 

C  Annuity,  .      } 

Given  <  Time  to  Run,  and  >  to  find  the  rate  of  in- 

(  Present  Value  of  an  Annuity,  )      terest. 

From  the  formula  (6),  Case  III,  £*=-& — ?X      pn    »  we  ODtain 

=-^r  ....  (1).  This  is  the  simplest  expression  we  can  ob- 
tain for  the  rate  as  the  equation  is  of  the  ^th  degree  and  can  not 
be  solved  in  a  general  manner. 

I.     If  an  immediate  annuity  of  $80,  running  14  yr.,   sells  for 
$650,  what  is  the  rate? 

By  formula, 

^n— 1_^__$650        _     or 


————^-—=8.125.     Solving  this   equation   by   the   method  of 

Double  Position,  we  find  r=S%-\-- 

By  another  method. 

$65Or-$80=8.125.     By  referring  to  a  table  of  present  worths 
of  $1,  corresponding  to  14  years,  we  find  it  to  be  between  8  and 


126  FINKEL'S   SOLUTION  BOOK. 


PROBLEMS. 

1.     What  is   the  amount  of  an   annuity    of  $1000,    forborne 
15  years,  at  3£%  compound  interest?  Ans.  $19295.125 


2.  What    will    an    annuity    of    $30    payable    semi-annually, 
amount  to,  in  arrears  3  years  at  7%  compound  interest? 

Ans.  - 

3.  What  is  the  present  worth  of  an   annuity   of  $500   to  con- 
tinue 40  years  at  7%  ?  Ans.  - 

4.  What  is  the  present  worth   of  an   annuity   of  $200,   for  7 
years,  at  5%  ?  Ans.  $1152.27. 

5.  A  father  presents  to  his  daughter,  for  8  years,   a   rental  of 
$70  per  annum,  payable  yearly,   and   the  reversion   for   12  years 
succeeding  to-  his  son.     What  is  the  present  value  of  the   gift  to 
his  son,  allowing  4%  compound  interest?  Ans.  - 

6.  A  yearly  pension  which  has  been  forborne  for   6   years,  at 
6%,  amounts  to  $279  ;  what  was  the  pension?  Ans.  $480.03. 

7.  A  perpetual  annuity  of  $100  a  year   is   sold   for   $2000  ;  at 
what  rate  is  the  interest  reckoned?  Ans.  - 

8.  A  perpetual  annuity  of  $1000  beginning  at  the   end   of  10 
years,  is  to  be  purchased.     If  interest  is  reckoned   at  3-J%,  what 
should  be  paid  for  it?  Ans.  -  — 

9.  If  a  clergyman's  salary  of  $700  per  annum  is  6  years  in  ar- 
rears, how  much  is  due,  allowing  compound  interest  at  6%  ? 

Ans.  $4882.72. 

10.  A  soldier's  pension  of  $350  per  annum   is   5   years   in  ar- 
rears; allowing  5%  compound  interest,  what  is  due  him? 

Ans.  $1933.97. 

11.  What  annual  payment  will  meet  principal  and  interest  of 
a  debt  of  $2000  due  in  4  year  a  8%  compound  interest?     Ans.  — 

12.  What  is  the  present  worth  of  a  perpetual  annuity  of  $600 
at  6%  per  annum?  Ans.  $10000. 

13.  What  is  the  present  value  of  an  annuity  of  $1000,  to  com- 
mence at  the  end  of  15  years,  and   continue   forever,   at   6%  per 
annum?  Ans.  $6954.40. 

14.  To  what  sum  will  an  annuity  of  $120  for  20  years  amount 
at  6%  per  annum?  Ans.  $4414.27. 

15.  A  debt  of  $8000  at  6%  compound  interest,   is  discharged 
by  eight  equal  annual  installments;  what  was  the  annual  install- 
ment? Ans.  $1288.286- 


MISCELLANEOUS   PROBLEMS. 


127 


CHAPTER  XV. 

MISCELLANEOUS  PROBLEMS, 

INVOLVING    THE    VARIOUS    APPLICATIONS    OF    PERCENTAGE. 

I.     Sold  a  cow  for  $25,  losing  16f%  ;  bought  another  and  sold 


it  at  a  gain  of  16% ;  I  neither  gained  nor  lost  on   the  two  ;  what 
Was  the  cost  of  each? 

-1.   100%— cost  of  the  first  cow. 

2.  16f  %— loss. 

3.  100%— 16|%— 83^%— selling  price. 

4.  $25— selling  price. 

5.  .-.  83-J%- $25, 


A. 


B. 


=-     of  $25— $.30,  and 


in. 


6. 


7.  100%— 100  times  $.30— $30=cost  of  first  cow. 

8.  $30 — $25— $5,  loss  on   the  first  cow,   and  gain  on 

second  cow. 

1.  100%— cost  of  second  cow. 

2.  16%— gain. 

3.  $5— gain. 

4.  .-.  16%— $5. 

5.  1%— TV  of  $5— $.3125,  and  [cow. 

6.  100%— 100  times  $.3125— $31.25— cost   of  second 
(  $30=cost  of  first  cow,  and 

)  $31.25=cost  of  second  cow. 


Remark. — Since  I  lost  $5  on  the  first  cow,  and  neither  gained 
»**r  lost  on  the  two,  I  must  have  gained  $5  on  the  second  cow. 
.•  16%— $5. 

I.     There  have  been  two  equal  annual  payments  on  a  6%  note 
of  $175,  given  2   years   ago   this   day0     The  balance   is 
L40  ;  what  was  each  payment? 


II. 


(1.)         100  %=a  payment. 
(2.)        100%— $175, 
(3.)        l%=ri(5-  of  $175=$1.75,  and 
(4.)         6%=  6  times  $1.75=$10.50=interest  for  1  year. 
(5.)         $175+$10.50=  $185.50 —  amount  before  paying 
the  payment.  [payment. 

(6.)         $185.50 — 100%  —amount    left    after    paying    the 

1.  100%— $185.50— 100%, 

2.  1%— Tio    of  ($185.50— 100%)— $1.855— 1%,  and 

3.  6%—  6  times    ($1.855— 6%)— $11.13— 6%— inter- 
(7.)J  est  for  second  year. 

4.  $185.50— 100%+$11.13—6%— $196.63— 106%  =» 

amount  before  paying  the  last  payment. 

5.  $196.63  —  106%  —  100%  —  $196.63  —  206%  = 

amount  left  after  paying  the  last  payment. 


128 


FINKEL'S   SOLUTION   BOOK. 


(8. 

(9. 
(10.) 
(11.) 
(12.) 


$154.40=amount  after  paying  the   last   payment 

.-.  $154.40=$196.63— 206%. 

206%— $196.63— $154.40— $42.23, 

1%— ^  of  $42.23=$.205,  and 

100%— 100  times  $.205— $20.50=the  payment. 


III.     .*.  |20.50=the  payment. 

Remark. — In  this  solution  we  are  obliged  to  use  the  minus 
sign,  — ,  which  is  no  obstacle  to  the  student  of  algebra,  but  to  the 
student  of  arithmetic  it  may  seem  insurmountable.  To  avoid 
this  sign,  we  give  another  solution. 


II. 


III. 


(I-') 

(2.) 

(3.) 
(4-) 
(5.) 
(6.) 

(7.) 
(8.) 


(9.) 


(10,) 


(11.) 
(12.) 
(13.) 
(14.) 
(15.) 


100%— the  payment.     Then 

$154.40+100%— amount  of  the  debt  at  the  end  of 

of  the  second  year. 

100%— principal  that  produced  this  amount. 
6  %— interest. 
106%—  amount. 

...  106%— $154.40+100%,  [and 

l%-Ti¥  of  ($154.40+100%  )-$1.4566^+M%, 
100%— 100  times  ($1.4566^+11% )  =  $145.66^ 

-|-94i|%— amount  at  end  of  the  first  year  after 

paying  off  the  payment. 
$145.66^+94/3  %+100%—  $145.66^  +  194|f % 

—amount   before    paying    oft'  the    payment  — 

amount  at  end  of  first  year. 
100%— the  principal  that  produced  it. 
6%=interest. 
106%— amount. 


l-83^fr%,  and 
100%  =  100  times   ($1.3?Hw  + 

$137JtfH-183^/V%=tibe  amount  at  first. 
$175=the  amount  at  first. 


and 
100%=100  times  $.205—  $20.50=the  payment. 


.'.  $20.50—  the  payment. 


(R.  H.  A.,  p.  26^  prob.  5.) 


Explanation.  —  $154.40=the  amount  after  paying  off  the  last 
payment.  .-.  $154.40-)-100%=amount  before  paying  of  the  last 
payment,  or  it  equals  the  debt  at  the  end  of  the  first  year  plus 
the  interest  on  this  debt  for  the  second  year.  .-.  We  let  100%= 
the  debt  at  the  end  of  the  first  year,  106%—  amount  of  100%  for 
1  year.  /,  106%  =  $154.40  +  100%.  Then  proceed  as  in  the 
solution. 

I.     If  a  merchant  sells  f  of  an   article    for   what  J   of  it  cost, 
what  is  his  gain  %  ? 


MISCELLANEOUS   PROBLEMS. 


129 


1.  100%=cost  of  whole  article. 

2.  87i%=J  of  100%=cost  of  £  of  the  article. 

3.  87^-%=selling  price  of  f  of  the  article. 

4.  29i%=£  of  87|%=selling  price  of  £  of  the  article. 

5.  116f%=4    times    29^%=  selling    price    of  the    whole 

article. 

6.  .'.  116|%— 100%=16f%=gain. 

.-.  16|%=his  gain.  (Milne's  Prac.,  p.  360,prob.  51.) 

A  merchant  sold  goods  to  a  certain  amount,  on.  a  commis- 
sion of  4%,  and  having  remitted  the  net  proceeds  to  the 
owner,  received  ^%  for.  prompt  payment,  which 
amounted  to  $15.60.  What  was  his  commission? 

100%=cost  of  goods. 

4%=commission. 

100%— 4%=96%=net  proceeds. 

;!%— amount  received  for  prompt  payment. 

2.  $15.60=amount  received  for  prompt  payment. 

3.  .-.  i%=$15.60. 

4.  1%=4  times  $15.60=$62.40. 

5.  100%  =100  times  $62.40=$6240=net  proceeds. 
.-.  96%=$6240. 

1%=^  of  $6240=$65,  and 

100% =100  times  $65=$6500=cost  of  goods. 

100%  =$6500. 

1%=^  of  $6500=$65,  and 

4%=4  times  $65=$260=his  commission. 

His  commission=$260. 

(  Greenleafs  N.  A.,  p.  J^l.prob.  11.) 


(2.) 
(3.) 


(4.) 


(5.) 
(6.) 
(7-) 

(8.) 


If  I  sell  30  yards  of  cloth  for  $132,  and  gain  10%,  how 
ought  I  to  sell  it  a  yard  to  lose  25%  ? 


$132=selling  price  of  30  yards. 
$4.40=$132-^30=selling  price  of  one  yard. 
100  %  —cost  of  one  yard. 
10%=gain. 

100%+10%=110%=selling  price  per  yard. 
$4.40=selling  price  per  yard. 
.-.  110%=$4.40. 
1  %=TH  of  $4.40=$.04, 
100%=100  times  $.04=$4=cost  per  yard. 
I.  100%=$4. 


HI. 


(1.) 

(2.) 

(  3.  ) 

(4.) 

(5.1 

(6  ) 

(7. 

(8. 

(9.) 


3.  25%  =25  times  $.04=$l=loss. 

4.  ...  $4  —  ^i—^^selling  price  per  yard  to  lose  25%. 

.•.  I  must  sell  it  at  $3  per  yard  to  lose  25%. 

(StoddarcCs  Complete,  p.  206,  prot>.  9.) 


130  FINKEL'S  SOLUTION  BOOK. 

I.  A  merchant  receives  on  commission  three  kinds  of  flour  ; 
from  A  he  receives  20  barrels,  from  B  25  barrels,  and 
from  C  40  barrels.  He  finds  that  A's  flour  is  10%  better 
than  B's,  and  that  B's  is  20%  better  than  C's.  He  sells 
the  whole  at  $6  per  barrel.  What  in  justice  should 
each  man  receive? 

(1.)  $6=selling  price  of  1  barrel. 

2.)  $510=sellmg  price  of  (20+25+40),  or  85  barrels. 

3.)  100%—value  of  C's  flour  per  barrel. 

(4.)  120%—value  of  B's  flour  per  barrel. 

(1.  100%=120%. 

(5.W2.  1%=^  ofl20%=li%, 

13.  10%— 10  times  1-J %=12%. 

(6. )  120%  +12  %— 132  %=valtie  of  A's  flour  per  barrel. 


II. 


(7.)         4000%=40  times  100%— what  C  received. 


i! 


(8.)         3000%— 25  times  120%— what  B  received. 


2640%— 20  times  132%— what  A  received. 
9640%— 4000%+3000%+2640%— what  all  rec'd. 
$5 10=  what  all  received. 
9640^=4510. 


(9. 

(10. 
(11. 
(12. 

(13.)  1%=^  of  $510— $.52|-J|,  and  [received. 

(14.)  4000%— 4000  times  $.52fii  — $21H||=what  C 

(15.)  3000%— 3000  times   $. 52-Jii  —  $158^1— what  B 

received.  [received. 

(16.)  2640%— 2640  times    $.52fi|  — $139iJ|— what  A 


v  Tic»i/2"^y— A's  share, 
ill.  .'.  I  $158.Hf— B's  share,  and 
:C's  share. 

(  Greentcafs  National  Aritk.  p.  442.) 

I.  f  of  B's  money  equals  A's  money.  What  %  is  A's  money 
less  than  B's,  and  what  %  is  B's  money  more  than  A's? 

1.  100%— B's  money. 

2.  75%— |  of  100%— A's  money. 

TT  j  3.  100%— 75%— 25%=excess  of  B's  money  over  A's. 
11X4.  75%— 100%  of  itself, 

5.  1%— TV  of  100%— H%,  and  [than  A's. 

6.  25%— 25  times  l£%=33i%:=the  %  B's  money    is  more 

A's  money  is  25%  less  than  B's,  and 

B's  money  is  33£%  more  than  A's  money. 

(Stod.  Comp.,p.  203,prob.  19.) 

I.  At  what  price  must  an  article  which  cost  30  cents  be 
marked,  to  allow  a  discount  of  12^%  and  yield  a  net 
profit  of  16f%  ? 


MISCELLANEOUS   PROBLEMS. 


131 


II. 


1. 

2. 
3. 

(4. 


100%=30/, 

16f%=16£  times  ^Xxm5/=profit. 

30/+5/=35/=the  price  at  which  it  must  sell  to 


III. 


I. 


1.  100%=marked  price. 

2.  12^%=discount  from  marked  price. 

3.  100%— 12|%=87i%=selling  price. 

4.  35/=selling  price. 
(5.n  5.  .-.  87|%=35/. 

6.  l%=^~  of  35/=.40/,  and 

7.  100%=100  times  .40/=40/=marked  price. 
/.  40^=marked  price. 

(Seymour's  Prac.,  p.  203; prob.  4.) 

Had  an  article  cost  10%  less,  the  number  of  %  gain  would 
have  been  15%  more  ;  what  was  the  gain? 

1.  .7#6>%=selling  price. 

2.  100%— actual  cost  price. 

3.  100%—  100%=gain. 

4.  100%  — 10%=90%=supposed  cost. 

5.  100% — 90%=conditional  gain. 

6.  90%=100%  of  itself. 


[difference. 


8.  100%—  90%=( 

=conditional  gain  %. 

9.  /.  V  X100%— 100%—  (100%  — 

10.  15%=difference. 

11.  /.  ix^#%=15%.  [the  actual  cost. 

12.  100%=9  times  15%=135%=selling  price  in   terms  of 

13.  .-.  135%— 100%=35%=gain. 

.*.  35%=gain.  (R.  H.  A.,  p.  406,prob.  87.) 

A  literal  solution. 
Let  5=:selling  price  and  C=the  cost.     Then  5 — C=gain  and 

C* S~*  C1 

• — -~r-  =rate  of  gain.     S — T9^C=conditional  gain   and 

^  10  $ Q        $ Q 

-=conditional  rate   of  gain.     .•.  — — ^___^.^  or 

.'.  1.35C—C=.35C=gain. 


III. 


whence  5=fJ  C=1.35  C. 
.-.  Rate  of  gain=.35C-7-C==.35=35%. 

In  the  erection  of  my  house  I  paid  three  times  as  much  for 
material  as  for  labor.  Had  I  paid  6%  more  for  labor, 
and  10%  more  for  material,  my  house  would  have  cost 
$3052.  What  did  it  cost  me? 


132  FINKEL'S   SOLUTION  BOOK. 

(1.)  100%=cost  of  labor. 

(2.)  300%— 3  times  100%=cost  of  material. 

rl.  100%=100%,     i 

,„  J2.  !%=!%,  and 

<6-')3.  6%=6%. 

U.  100%+6%=106%=supposed  cost  of  labor. 

II.  100%  =300%, 

2.  l%=Ti-o-  of300%=3%,  and 

3.  10%=10  times  3%=30%. 

4.  300%+30%=330%=supposed  cost  of  material. 

(5.)  330%+106%==436%=:SUPPosed  cost  of  house. 

(6.)  $3052=supposed  cost  of  house. 

(7.)  .-.436%  =$3052, 

(8.)  l%=^k  of  $3052=$7,  and 

(9.)  100%=100  times  $7=$700=cost  of  labor. 

(10.)  300%=300  times  $7=$2100=cost  of  material. 

( 11.)  $2100+$700=$2800=cost  of  house. 

III.   .-.  $2800=cost  of  the  house. 


I.  I  invest  |-  as  much  in  8%  canal  stock  at  104%,  as  in  6% 
gas  stock  at  117%  ;  if  my  income  from  both  is  $1200, 
how  much  did  I  pay  for  each,  and  what  was  the  income 
from  each  ? 

(1.)  100%=investment  in  gas  stock.     Then 

(2.)  66f  %=investment  in  canal  stock. 

1.  100%=par  value  of  the  gas  stock. 

2.  117%=market  value  of  the  gas  stock. 

3.  .-.  117%=100%,from  (1), 

4.  1%=^-^  of  100%=-fr5-%  >  and 

5.  100%=100    times    }fo%=85||%=par   value    in 

terms  of  the  investment. 
;i.  I00%=85|f%, 
(5.)<^2.  l%=ff-?-%,  and 

U.  6%=6  times  |^%=5A%=income  of  gas  stock, 

II.  100%=par  value  of  canal  stock. 
2.   104%=market  value. 
3.  /.  104%=66|%, 
4.  l%=T^of66|%=||%,and 
100%=100  times 


(3.) 


(6.K2.  l%=zT^  of  64^%=|f%,  and 

13.  8%=8  times  ff  %=5^%==mcome  of  canal  stock. 
(7.)         5^g-%-}-5^=10^%=income  from  both. 
(8.)         $1200=income  from  both. 
(9.)         ,. 


(10.)         1%=-—-  of  $1200=$117,  and 


MISCELLANEOUS   PROBLEMS.  133 

(11.)         100%=100    times    $117=$11700=investment    in 

gas  stock.  [canal  stock. 

(12.)         66f%=66f   times    $117=$7800  =  investment    in 

(13.)         5A%=5A  times  $117=$600=income  from  each. 


$600=income  from  each. 

JJf       .'.      $11700=investment  in  gas  stock,  and 
$7800=investment  in  canal  stock. 

I,     A  man  bought  two  horses  for  $300;  he  sold  them  for  $250 
apiece.     The   gain   on   one  was  5%   more  than  on  the 
.  other;  what  was  the  gain  on  each? 

1.  $300=cost  of  both. 

2.  $500=$250+$250=selling  price  of  both. 

3.  $500—  $300=$200=gain  on  both. 

4.  100%=gain  on  first  horse.     Then 

5.  105%=gain  on  second  horse. 

n<|   6.  205%=100%+105%=gain  on  both. 

7.  $200=gain  on  both. 

8.  .-.  205%=$200. 

9.  l%=?fa  of  $200=$Jf,  and 

10.  100%=100  times  $|f=$97.56¥4T=gain  on  the  first. 

11.  105%=105  times  $££=$102.43f£=gain  on  the  second. 

Ill       •    $  $97.56^T=gain  on  the  first,  and 
'   (  $102.43f|=gain  on  the  second. 

Note.  —  In  this  solution,  it  is  assumed  that  the  gain  on  one  was 
5%  of  the  gain  on  the  other  more  than  the  other,  and  this  is  the 
usual  assumption.  But  the  problem  really  means  that  the  per 
cent,  of  gain  on  one,  computed  on  its  cost,  was  5%  more  than 
the  per  cent,  of  gain  on  the  other,  computed  on  its  cost.  By  this 
assumption,  the  problem  is  algebraic.  The  following  is  the 
solution:  Let  #=the  cost  of  the  first  horse,  and  $300  —  #,  the 
cost  of  the  second.  Then  $250  —  #—  gain  on  first,  and  $250  — 
($300—  x)=x—  $50,  the  gain  on  the  second.  ($250—  x)  -t-x= 
rate  of  gain  on  the  first,  and  (x  —  $50)-f-($300—  x),  the  rate  of 
gain  on  the  second.  Then  (250  —  x)-$-x  —  (x  —  50)-r-(300—  x)= 
j1^.  Whence,  by  clearing  of  fractions,  transposing  and,  combin- 
ing, *2__  10300  *=—i500000,  ^=5150^50^10009=  $147.7755, 
the  cost  of  the  first  horse.  $300—  *=$152.2245,  the  cost  of  the 
second  horse.  $250  —  #=$102.2245,  gain  on  the  first  horse,  and 
*  —  $50=$97.7755,  the  gain  on  the  second  horse. 

I.  An  agent  sells  produce  at  2%  commission,  invests  the 
proceeds  in  flour  at  3%  commission;  his  whole  commis- 
sion was  $75.  How  many  barrels  of  flour  did  he  buy 
at  $5  per  barrel  ? 


134 


FINKEL'S   SOLUTION  BOOK. 


II. 


III. 
I. 


II 


(1.)  100%=value  of  the  produce. 

(2.)  2 %— the  commission.  [vested  in  the  flour, 

(3.)  100% — 2%=98%=net  proceeds,  or  amount  in- 

1.  100%=cost  of  the  flour. 

2.  3%=commission  on  flour. 

3.  100%+3%=103%=whole  cost  of  the  flour. 


(4.) 


(5.) 
(6.) 
(7.) 
(8.) 
(9.) 
(10.) 

(11.) 
(12.) 


.-.  103%=, 
l%=rta  of  98%=!%%,  and 
lOO%==100XT9-o8*%  =  95^%  =  cost  of  flour  in 
terms  of  the  value  of  the  produce. 
—  95TV\%=2T8o%%=commission  on  flour. 

%==whole  commission. 
«$75=whole  commission. 


l%^$75-MT8oV=  $15.45,  and  [produce. 

100%=100    times   $15.45=$1545  =  value   of  the 
95TV%%==95TVV  times   $15.45=  $1470  =  value  of 

the  the  flour. 
$5=costofl  barrel. 
$1470=cost  of  1470-=-5,  01  294  barrels. 


\ 
2. 

(3.) 
(4.) 


.*.  The  agent  bought  294  barrels  of  flour. 

A  distiller  sold  his  whisky,  losing  4%  ;  keeping  $18  of 
the  proceeds,  he  gave  the  remainder  to  an  agent  to  buy 
rye  at  8%  commission;  he  lost  in  all  $32  ;  what  was  the 
whisky  worth? 

(1.)         100%=value  of  the  whisky. 
>.)        4%=loss. 

100% — 4%=96%=amount  he  had  left. 
96% — $18=amount  he  invested  in  rye. 

1.  100%=cost  of  the  rye. 

2.  8%=commission  on  the  rye. 

3.  100%  +8 %=108%=whole  cost  of  rye. 

4.  .-.  108%=96%— $18, 

5.  1%=^  of  (96%— $18)=|%— $.16|,  and 

6.  100%=100    times    (f%  —  $.16f)= 88f%—  $16.66f 

=cost  of  rye. 

7.  8%=8    times   (|%— $.16f)=7i%— $1.33i=com- 

mission  on  rve. 

— $1.33^-)=!  H% — $1.33|=wholeloss. 
2=\vhole  loss. 
H^%_$1.33^=:$32 
*o= $33.33i, 

of  $33.33ih=$3,  and 


(5.) 


(6.) 

(7.) 
(8.) 
(9.) 

(10.) 
l(ll-) 


100%-=100  times  $3=$300=value  of  the  whisky. 


III.     .-.  $300=value  of  the  whisky. 


(Jt.  H.  A.,  p.  4Q6,  prob  91.) 


MISCELLANEOUS   PROBLEMS.  135 

I.  What  will  be  the  cost  in  New  Orleans  of  a  draft  on  New 
York,  payable  60  days  after  sight,  for  $5000,  exchange 
being  at  \\°/o  premium? 

1.  100%=face  of  the  draft. 

2.  l|%=premium. 

3.  100%+1-J%= 101i%=rate  of  exchange. 

4.  5%=discount  for  one  year. 


II. 


of  5%=discount  for  63  days. 


/.  101-4-%—  f  %=100|%=cost  of  the  draft 

7.  100%  =$5000. 

8.  1%=^  of  $5000=$50,  and 

lOOf  %=100f  times  $50=$5031.25=cost  of  the  draft. 

III.     .-.  $5031.25=cost  of  the  draft. 

Explanation. — Observe  that  since  the  draft  is  not  to  be  paid  in 
New  York  for  63  days,  the  banker  in  New  Orleans,  who  has  the 
use  of  the  money  for  that  time  allows  the  drawer  discount  on  the 
face  of  the  draft  for  that  time,  which  goes  (1)  towards  reducing 
the  premium  if  there  be  any,  and  (2)  towards  reducing  the  face 
of  the  draft. 

Note. — The  rate  of  exchange  between  two  places  or  countries 
depends  upon  the  course  of  trade.  Suppose  the  trade  between 
New  York  and  New  Orleans  is  such  that  New  York  owes  New 
Orleans  $10,250,000  and  New  Orleans  owes  New  York  $13,000,- 
000.  There  is  a  "balance  of  trade"  of  $2,750,000  against  New 
Orleans  and  in  favor  of  New  York.  Hence,  the  demand  in  New 
Orleans  for  drafts  on  New  York  is  greater  than  the  demand  in 
New  York  for  drafts  on  New  Orleans  and,  therefore,  the  drafts 
are  at  a  premium  in  New  Orleans.  But  if  New  York  owes  New 
Orleans  $13,000,000  and  New  Orleans  owes  New  York  $10,250,- 
000,  the  "balance  of  trade,"  $2,750,000,  is  against  New  York  and 
in  favor  of  New  Orleans.  Hence,  the  demand  in  New  Orleans 
for  drafts  on  New  York  is  less  than  the  demand  in  New  York 
for  drafts  on  New  Orleans  and,  therefore,  the  drafts  are  at  a  dis- 
count in  New  Orleans. 

If  the  trade  between  the  two  places  is  the  same,  the  rate  of  ex- 
change is  at  par. 

The  reason  why  the  banks  in  New  York  should  charge  a  pre- 
mium, when  the  balance  of  trade  is  against  them,  is  that  they 
must  be  at  the  expense  of  actually  sending  money  to  the  New 
Orleans  banks  or  be  charged  interest  on  their  unpaid  balance ; 
the  reason  why  the  New  Orleans  banks  will  sell  at  a  discount  is 
that  they  are  willing  to  sell  for  less  than  the  face  of  a  draft  in 
order  to  get  the  money  owed  them  in  New  York  immediately. 

Exchange  is  charged  from  -J  to  •£%,  and  is  designed  to  cover 
the  cost  of  transporting  the  funds  from  one  place  to  another. 


136 


FTNKEL'S   SOLUTION  BOOK. 


I. 


II. 


III. 
I. 


!!. 


What  will  a  30  days'  draft  on  New  Orleans  for  $7216.85 
cost,  at  •§%  discount,  interest  6%  ? 

1.  100  %=face  of  draft. 

2.  -f%=discount. 

3.  100%—  f  %=99f  %=face  less  the  discount. 

4.  6%=bank  discount  for  1  year. 

5.  %v%~'ffir  °f  6%=bank  discount  for  33  days. 

6.  99f  % — -i^%=99¥3^%=cost  of  the  draft. 

7.  100%=47216.85, 

8.  l%=Tfo.  of  $7216.85=$72.1685,  and 

9.  99^ ^^99^.   times   $72.1685=$7150.094=cost  of   the 

draft. 

.-.  $7150.094=cost  of  the  draft. 

The  aggregate  face  value  of  two  notes  is  $761.70  and  each 
has  1  year  3  months  to  run;  one  of  the  notes  I  had  dis- 
counted at  10%  true  discount  and  the  other  at  10% 
bank  discount,  and  realized  from  both  notes  $671.50. 
Find  the  face  value  of  both  notes. 

100%=face  of  note  discounted  at  bank   discount. 
$761.70 — 100%=face  of  note   discounted  at  true 

discount. 

10%— bank  discount  for  1  year. 
12^%=bank  discount  for  1  year  3  months. 

1.  100%=present  worth  of  second  note. 

2.  10%=interest  on  present  worth  for  1  year. 

3.  12£%=interest  for  1  year  3  months. 

4.  100%+12|%=112i%=amount  of  present  worth. 

5.  $761.70 — 100%=amount  of  the  present  worth. 


(1.) 

(2.) 

(3.) 
(4.) 


(5.) 


(6.) 

(70 

(8.) 

(9.) 

(10.) 

(11.) 
(12.) 

(13.) 


6.  .-.  112i%=$761.70— 100%, 


of    ($761.70— 100%  )=:$6.7706f— 1%, 


100%=100  times  ($6.7706|  — 1%)  =  $677.06|— 

88f  %=present  worth. 
$761.70— 100%— ($677.06|— 88f  % )  =  $84.63-J  — 

H^.%— true   discount.  [discount. 

$84.63i— ll^-%t+12i%=$84.63i+  1TV  %  =  whole 
$761.70— $671.50=$90.20=whole  discount. 


l%=jrof  $5.56|=$4.008,  and 

100%=100   times  $4.008=$400.80=face  of  note 

discounted  at  bank  discount. 
$761.70—  100%==$761.70—  $400.80=$360.90=face 

of  note  discounted  at  true  discount. 


III. 


.    (  $400.80=face  of  note  discounted  at  bank  discount,  and 
(  $360.90=face  of  note  discounted  at  true  discount. 


MISCELLANEOUS   PROBLEMS. 


137 


II. 


II. 


3.) 

(*•)• 

(5.) 
(6.) 
(V.) 

19 

10.) 


1.      A  merchant  bold  part  of  his  goods  at  a  profit  of  20%,  and 
the  remainder  at  a  loss  of  11%.       His  goods  cost  $1000 
and  his  gain  was  $100;  how  mnch  was  sold  at  a  profit? 
100%— cost  of  goods  sold  at  a  profit.      Then 
$1000 — 100%— cost  of  goods  sold  at  a  loss. 
20%— profit  on  100%,  the  part  sold  at  a  profit 

1.  100%— $1000— 100%. 

2.  1%—  -j-^of  ($1000— 100%)— $10— 1%, 

3.  11%— 11  times  ($10— 1%)— $110— 11%— loss  on 

the  remainder. 

.'.  20%— ($110— 11%)—  31%—  $110=gain. 
$100— gain. 
/.  31%— $110— $100. 
31%— $210, 

1  %— ^T  of  $210—  $6ff ,  [profit. 

100%— 100  times  $6ff— 677.41ff— part  sold  at  a 
III.     .-.  $677.41f-f— value  of  the  part  sold  at  a  profit. 

I.  By  discounting  a  note  at  20%  per  annum,  I  get  22-J% 
per  annum  interest;  how  long  does  the  note? 

1.  22-£%  of  the  proceeds=20%  of  the  face  of  the  note. 

2.  1%  of  the  proceeds— —f  of  20%=|%  of  the  face  of  the 

note.  «** 

3.  100%  of  the  proceeds=100  timss  |%=88|%  of  the  face 

of  the  note. 

4.  100%— face  of  the  note. 

5.  88f%— proceeds. 

6.  100% — 88f  %=ll£%=discount  for  a  certain  time. 

7.  20%— discount  for  360  days. 

8.  1%— discount  for  -fa  of  360  days,  or  18  days. 

9    lli%==discount  for  11^  times  18  days,  or  200  days. 
III.     The  note  was  discounted  for  200  days. 

I.  A  man  bought  a  farm  for  $5000,  agreeing  to  pay  princi- 
pal and  interest  in  5  equal  annual  installments.  What 
will  be  the  annual  payment  including  interest  at  6%? 

1.  100  c/c=on&  annual  payment. 

2.  .-.  100%=amount  paid  at  end  of  the  fifth  year 

since  the  debt  was  then  discharged. 

3.  100  %-principal  that  drew  interest  the  fifth  year. 

4.  6  %= interest  on  this  principal. 

5.  .*.  KXH+6%=106%=amount  of  this  principal. 

6 .  . ' .  1 06  #F=100  %  -  the  annual  payment . 

7.  l  %=^  of  100  #=£ f  %,  and 

8.  100^=100  times  f|%=94£f  %=  principal  at   the 

beginning  of  the  fifth  year. 

9.  94J|  %-hlOO  $>=194jf  ^-amount  at  the  end  of  the 

fourth  year. 


II 


138 


FINKEL'S  SOLUTION  BOOK. 


II. 


(2.) 


(3.) 


'1.   100%=principal  at  the  beginning  of  the   fourth 
year. 

2.  6%=interest  on  this  principal. 

3.  100%+6%=106%=amount. 

4.  .".  106%=194££%, 

5.  1%=T^  of  194ff%==1.83Y9¥5-^g-%,  and 

6.  100%=100  times  1.83^r%%=183^\3¥%=princi- 

pal  at  the  beginning  of  the  fourth  year. 

7.  lS&f^%+lQO%=285Jffo%==amount  at  the  end 

of  the  third  year. 

1.  100%=principal   at  the   beginning   of  the   third 

year. 

2.  6%=interest.  [third  year. 

3.  100%+6%=106%  =  amount  at  the  end  of  the 

4.  .-.  106%=283^ftftF%, 

6.  100%3oO  time^F67TVAVT%?^67T4^rV%  = 

principal  at  the  beginning  of  third  year. 

7.  267AWTV%+100%— 367T4A88rV%^  amount  at 

the  end  of  second  year. 

1.  100%=principal  at  the  beginning  of  second  year. 

2.  6%=interest  [year. 

3.  100%+6%=106%=amount  at  the  end  of  second 

4.  .-.  106%=367T\\8-84rr%> 

6.  100%=100  times  3.46fff|m%=346|fff|-|-f  %= 
principal  at  the  beginning  of  the  second  year. 

the  end  of  first  year. 

1.  100%— principal    at    the    beginning    of   the   first 

year,  or  the  cost  of  farm. 

2.  6%=  interest. 

3.  100%+6%=106%=amount  at  end   of   first  year. 

4.  . 

5.  1 

6.  100%=100  times  4.2; 

%=cost  of  the  farm. 
(6.)         $5000=cost  of  the  farm. 

(8.)         l%=$5000^421^\^%44%Vz:=$ll-8698-|-,  and 
(9.)         100%=100   times  $11.8698=$1186.98+=the  an- 
nual payment. 

III.     .'.  $1186.98+ =the  annual  payment. 

(Milne's  Prac., p.  361, prob.  63.) 


(4.) 


(5.) 


I.     A  and   B   have  $4700  ;  f 

~i% 

B's  share;  how  much  has  each? 


of  A's  share  equals 


of 


PROPORTION. 


II. 


of  B's, 


8.  1%  of  A's=    -  of  !%%=%%%  of  B's,  and 

9.  100%  of  A's=100  times  fo%=74X%  of  B's. 

10.  100%—  B's  share. 

11.  74^2T%=A's  share. 

12.  100%+7422T%=17422T%==sum  of  their  shares. 

13.  $4700=sum  of  their  shares. 

14.  .-.  174  22T%  =$4700, 


15".  l== 


>f$4700==$27,  and 

16.  lOOftf^lOO  times  $27=$2700=B's  share. 

17.  74^2T%=74Y2T  times  $27— $2000=A's  share. 
(  $2700=B's  share     and 

'    "I  $2000=A's  share. 


1.     J 


CHAPTER    XVI. 

RATIO  AND    PROPORTION. 


1.     Rdtio  is  the  relative  magnitude  of  one  quantity  as  com 
pared  with  another  of  the  same  kind;  thus,  the  ratio  of  12  apples 
to  4  apples  is  3. 

The  first  quanity,  12  apples,  is  called  the  Antecedent,  and  the 
second  quantity,  4  apples,  the  consequent.  Taken  together  they 
are  called  Terms  of  the  ratio,  or  a.  Couplet. 


the  common  sign  of 


2.      The   Sign  of  ratio  is  the  colon, 
division  with  the  horizontal  line  omitted. 

Note.  —  Olney  says,  "There  is  a  common  notion  among  us,  that  the  French 
express  a  ratio  by  divding  the  consequent  by  the  antecedent,  while  the  En- 
glish express  it  by  dividing  the  antecedent  by  the  consequent.  Such  is  not 
the  fact.  French,  German,  and  English  writers  agree  in  the  above  defini- 
tion. In  fact,  the  Germans  very  generally  use  the  sign  :  instead  of  -f-;  and 


140  FINKEL'S   SOLUTION   BOOK. 

by  all,  the  two  signs  are  used  as  exact  equivalents."  Some  writers, 
however,  divide  the  consequent  by  the  antecedent,  as  a  :  b— —  This  is  ac- 
cording to  Webster's  definition  and  illustration.  To  my  mind,  to  divide 
the  antecedent  by  the  consequent  is  more  simple  and  philosophical  and 
should  be  universally  adopted  by  all  writers  on  mathematics. 

3.  A  Direct  JKatio  is  the  quotient  of  the   antecedent  di- 
vided by  the  consequent. 

4.  An  Indirect  Hatio  is  the  quotient  of  the  consequent 
by  the  antecedent. 

5.  A  ratio  of  Greater  Inequality  is  a  ratio  greater  than 
unity;  as,  7:3. 

6.  A  ratio  of  Less  Inequality  is  a  ratio  less  than  unity; 
as,  4  :5. 

7.  A  Compound  Hatio  is  the  product  of  the  correspond- 
ing terms  of  several  simple   ratios.     Thus,  the  compound  ratio  of 
1 :  3,  5  :4,  and  7  :  2  is  1x5x7:  3X4X2. 

8.  A  Duplicate  Ratio  is  the  ratio  of  the  squares  of  two 
numbers. 

9.  A  Triplicate  Hatio  is  the  ratio  of  the  cubes  of  two 
numbers;  as,  a3  :   bz. 

1C.     A   SubduplicateJRatio  is  the    ratio  of  the  square 
roots  of  two  numbers;  as,  \/^:  \/~&. 

11.  A   Subtriplicate  Ratio  is  the  ratio  of  the  cube  roots 
of  two  numbers;  as,  $/^~:  $/£. 

PROPORTION. 

% 

12.  Proportion  is  an  equality  of  ratios.     The  equality  is 
indicated  by  the  ordinary  sign  of  equality  or  by  the  double  colon,  : :. 
Thus .  a  :  b=c  \d,  or  a  :  b  :  :  c  :  d. 

13.  The  Extremes  of  a  proportion  are  the  first  and  fourth 
terms. 

14.  The  Means  are  the  second  and  third  terms. 

15.  A  Mean  Proportional  between  two  quantities  is  a 
quantity  to  which  either  of  the  two  quantities  bears  the  same  ratio 
that  the  mean  does  to  the  other  of  the  two. 

16.  A    Continued  Proportion  is  a  succession  of  equal 
ratios,  in  which   each   consequent  is  the  antecedent  of  the  next 
ratio. 

17.  A     Compound    Proportion    is    an    expression    of 
equality  between  a  compound  and  a  simple  ratio. 


PROPORTION.  141 

A  Conjoined  Proportion  is  a  proportion  which 
has  each  antecedent  of  a  compound  ratio  equal  in  value  to  its 
consequent.  The  first  of  each  pair  of  equivalent  terms  is  an  an- 
tecedent, and  the  term  following,  a  consequent.  This  is  also 
called  the  "Chain  rule." 

What  is  the  ratio  of  -J  to  f  ? 
£-s-f  =  i  X  f  =  |,the  ratio. 

What  is  the  ratio  of  10 bu.  to  If  bu.? 
10  bu.  -i-  If  bu.  =  10  X  TV  =  7,  the  ratio. 

What  is  the  ratio  of  25  apples  to  75  boxes? 
Ans.     No  ratio  ;  for  no  number  of  times  one  will  produce 
the  other 

In  a  true  proportion,  we  must  always  have  greater  :  less  :: 
greater  :  less  or  less  :  greater  : :  less  :  greater.  The  test  for  the 
truth  of  a  proportion  is  that  the  product  of  the  means  equals  the 
product  of  the  extremes. 

I.  Ifa5-cent  loaf  weighs  7oz.  when  flour  is  $8  per  barrel, 
how  much  should  it  weigh  when  flour  is  $7.50  per  barrel? 

It  should  evidently  weigh  more. 

.•.  less  :    greater  : :  less  :  greater. 

$7.50  :  $8.00      : :  7oz  :  (?  =  7Ty>z.) 

I.  If  a  staff  3  feet  long,  casts  a  shadow  2  feet,  how  high  is 
the  steeple  whose  shadow  at  the  same  time  is  75  feet? 

Since  the  steeple  casts  a  longer  shadow  than  the  staff,  it  is  evi- 
dently higher  than  the  staff. 

.-.  less  :  greater  : :     less    :  greater. 

2  feet  :  75  feet    :  :  3  feet  :   (  ?=112|  feet.) 

I.     What    number  is  that  which   being   divided  by  one   more 
than  itself,  gives  -ij-  for  a  quotient? 


II.. 

III. 
I. 

1. 

2. 

3. 
4. 
5. 

6. 

7. 

M 

Let  f=number.     T 

hen 
.  :  :  1  :  7,  whence 

lumber, 
ed  by  3  more    than  itself  gives  $  for 

f+1      ^01^  +  - 
7(*)=1(*+1)  or 
Y*=f+l;  whence 

i=-rV»  and 
|=2  times  TL=i=i 

i=the  number. 

/"hat    number   divid 
a  quotient? 

142  FINKEL'S   SOLUTION  BOOK. 

1.   Let  f±=the  number.     Then 
2 

=|-  or,  putting  this  in  the  form  of  a  proportion, 


1-+3  : :  7  :  9.  [the  product  of  the  extremes. 

=  1^-j-21>  the  product  of  the   means  being  equal  to 
'—¥=4=21, 

6.  £=±of21=5i,  and 

7.  |=2  times  5i=10i=the  number. 
III.     .-.  10£=the  number. 

I.  If  7  lb.  of  coffee  is  equal  in  value  to  5  lb.  of  tea,  and  3  lb. 
of  tea  to  13  lb.  of  sugar,  39  lb.  of  sugar  to  24  lb.  of  rice, 
12  lb.  of  rice  to  7  lb.  of  butter,  8  lb.  of  butter  to  12  lb. 
of  cheese  ;  how  many  lb.  of  coffee  are  equal  in  value  to 
65  lb.  of  cheese? 

1.  7  lb.  of  coffee=5  lb.  of  tea, 

2.  3  lb.  of  tea=13  lb.  of  sugar, 

3.  39  lb.  of  sugar=24  lb.  of  rice, 

4.  12  lb.  of  rice=7  lb.  of  butter, 

'^5.  8  lb.  of  butter=12  lb.  of  cheese,  and 
6.  65  lb.  of  cheese=  ?=39  lb.  of  coffee, 

7X3X39X12X8X65 
''      5X13X24X7X12 
III.  ^  /.  65  lb.  of  cheese=39  lb.  of  coffee. 

I.     I  can  keep  10  horses  or  15  cows  on  my   farm  ;   how  many 

horses  can  I  keep  if  I  have  9  cows? 
15  cows  :  9  cows  : :  10  horses  :   ?=6  horses. 
10  horses — 6  horses=4  horses. 
.-.  I  can  keep  4  horses  with  the  9  cows. 

I.  If  2  oxen  or  3  cows  eat  one  ton  of  hay  in  60  days,  how 
long  will  it  last  4  oxen  and  5  cows? 

2  oxen  :  4  oxen  : :  3  cows  :   ?=6  cows. 

.-.  4  oxen  eat  as  mnch  as  6  cows.  If  a  ton  of  hay  last  3  cows 
60  days,  it  will  last  6  cows,  which  are  equal  to  4  oxen,  and  5 
cows,  or  11  cows,  not  so  long. 

.-.  11  cows  I  3  cows  : :  60  days  :    ?=17T3T  days. 

I.  If  24  men,  by  working  8  hours  a  day,  can,  in  18  days,  dig 
a  ditch  95  rods  long,  12  feet  wide  at  the  top,  10  feet  wide  at  the 
bottom,  and  9  feet  deep;  how  many  men,  in  24  days  of  12  hours  a 
day,  will  be  required  to  dig  a  ditch  380  rods  long,  9  feet  wide  at 
the  top,  5  feet  wide  at  the  bottom,  and  6  feet  deep? 

95  rods  :  380  rods 

24  days  :  18  days 

12  hours  :  8  hours     _nA  men  .    p__12 


12  feet 

10  feet 

9  feet 


9  feet 

5  feet 

6  feet 


PROPORTION.  143 

380X18X8X9X5X6X24 


95X24X12X12X10X9 


=12  men. 


A  Louisville  merchant  wishes  to  pay  $10000,  which  he 
owes  in  Berlin.  He  can  buy  a  bill  of  exchange  in  Louis- 
ville on  Berlin  at  the  rate  of  $.96  for  4  reichmarks  ;  or 
he  is  offered  a  circular  bill  through  London  and  Paris, 
brokerage  |%  at  each  place,  at  the  following  rates  : 
£1=$4.90=25.38  francs,  and  5  francs=4  reichmarks. 
What  does  he  gain  by  direct  exchange? 

1.  $.238=1  mark. 

2.  $10000=10000-:-.238=42016.807  marks. 

3.  $.24=1  mark,  since  this  is  the  rate  of  exchange. 

4.  .-.  $10084.033=42016.807  times  $.24=42016.807  marks 

=direct  exchange. 

5.  42016.807  marks=(  ?=$10165.38.) 

6.  $4.90=£1—  1%  of£l=£.99J. 

7.  JE1=.99|.  times  25.38  fr. 

8.  5  fr.=4  marks. 

42016.807X4.90X5  ,00  .       . 

Circular 


II 


10.  $10165.38— $10084.033=$81.35  =  gain    by   direct    ex- 
change. 

III.     .'.  $81.35=gain  by  direct  exchange. 

I.  A  wheel  has  35  cogs;  a  smaller  wheel  working  in  it,  26  cogs; 
in  how  many  revolutions  of  the  larger  wheel  will  the 
smaller  one  gain  10  revolutions? 

1.  35  cogs — 26  cogs=9  cogs=what  the  smaller  wheel  gains 

on  larger  in  1  revolution  of  larger  wheel. 

2.  26  cogs  passed  through  the   point  of  contact=l  revolu- 

tion of  smaller  wheel. 

3.  1  cog  passed  through   the   point  of  contact—-^  revolu- 

tion of  smaller  wheel. 

II. ^  4.  9  cogs  passed  through  the  point  of  contact=-^-  revolu- 
tion of  smaller  wheel. 
5.   .'.  In  1  revolution  of  larger  wheel   the   smaller   gains  -£% 

revolution  of  smaller  wheel. 

/.  2^-  revolution  gained  :  10  revolutions  gained  .' :  1 
revolution  of  larger  wheel  :  ?=28f  revolutions  of 
larger  wheel. 

III.     .-.  The  smaller  wheel  will  gain  10  revolutions  in  28-f  revo- 
lutions of  larger  wheel. 

By  analysis  and  proportion. 

26  cogs  passed    through   the  point   of  contact==l  revolution  of 
the  smaller  wheel. 


144  FINKEL'S   SOLUTION   BOOK. 

35  cogs  passed  through  the  point  of  contact=l  revolution  of" 
the  larger  wheel.  But  when  the  larger  wheel  has  made  1  revo- 
lution, 35  cogs  of  the  smaller  wheel  have  passed  through  the 
point  of  contact.  If  26  cogs  having  passed  through  the  point  of 
contact  make  1  revolution  of  the  smaller  wheel,  how  many  rev-* 
olutions  will  35  cogs  make? 

By  proportion,  26  cogs  :  35  cogs  : :  1  rev.  :    ?=l^9-g-rev. 

.-.  The  smaller  wheel  makes  1/g-  revolutions  while  the  largei 
wheel  makes  1  revolution.  .-.  The  smaller  gains  1^-  revolutions, 
— 1  revolutions^-  revolution.  If  the  smaller  wheel  gains  -fa 
revoluion  in  1  revolution  of  the  larger  wheel  to  gain  10  revolu- 
tions on  t|ie  larger  wheel,  the  larger  wheel  must  make  more  rev- 
olutions. /.  less  :  greater  : .:  less  :  greater. 

•fa  rev.  ;  10  rev.   ;  :  1  rev.  of  larger    ;    ?=28f  rev.  of  larger. 

I.  If  the  velocity  of  sound  be  1142  fee.t  per  second,  and  the 
number  of  pulsations  in  a  person  70  per  minute,  what 
is  the  distance  of  a  cloud,  if  20  pulsations  are  counted 
between  the  time  of  seeing  the  flash  and  hearing  the 
thunder? 

1.  1142  ft.=distance  sound  travels  in  1  second. 

2.  68520  ft=60Xll42  ft.=distance  sound  travels  in  1  min., 

or  the  time  of  70  pulsations. 

3.  .'.  If  it  travels  68520  feet  while  70  pulsations   are  count- 

ed, it  will   travel   not   so   far  while   20   pulsations  are 
counted. 

4.  .'.  greater  :  less  : :  greater  :  less.  [145  yd.  2-iJ-  ft. 

5.  70  pul.  :  20pul.  : :  68520  ft.  :   ?=19577|  ft.=3  mi.  5  fur. 

III.     .-.  The  cloud  is  3  mi.  5  fur.  145  yd.  2|  ft.  distant. 

(/?.,  3d  p.,  p.  289,prob.  45.) 

PROBLEMS. 

1.  If  3  horses,  in  ^  of  a  month  eat  f  of  a  tori  of  hay,  how  long 
will  •§•  of  a  ton  last  5  horses? 

2.  If  a  4-cent  loaf  weighs  9  oz.  when  flour  is  $6  a  barrel,  how 
much  ought  a  5-cent  loaf  weigh  when  flour  is  $8  per  barrel? 

3.  A  dog  is  chasing  a  hare,  which  is  46  rods  ahead  of  the  dog. 
The  dog  runs  19  rods  while  the  hare  runs  17;  how   far   must  the 
dog  run  before  he  catches  the  hare? 

4.  If  52  men  can  dig  a  trench  355  feet  long,  60  feet  wide,  and 
8  feet  deep  in  15  days,  how  long  will  a  trench  be  that  is  45  feet 
wide  and  10  feet  deep,  which  45  men  can  dig  in  25  days? 

5.  If  -J-  of  12  be  3  what  will  £  of  40  be  ?  Ans.  15. 

6.  If  3  be  \  of  12,  what  will  £  of  40  be?  Ans.  6& 


II. 


PROBLEMS.  145 

If  18  men  or  20  women  do  a  work  in  9  days,  in  what  time 
•can  4  men  and  9  women  do  the  same  work?        Ans.  13^7T  days. 


8.  If  5  oxen  or  7  cows  eat  3T4T  tons  of  hay  in  87  days,  in  what 
time  will  2  oxen  and  3  cows  eat  the  same  quantity  of  hay? 

Ans.  105  days. 

9.  Divide  $600  between  three  men,   so   that   the   second  man 
shall  receive  one-third  more  than  the  first,  and  the  third  f  more 
than  the  second. 

10.  Two  men  in  Boston  hire  a  carriage  for  $25,  to  go  to  Con- 
cord, N.  H.,  and   back,   the   distance   being   72   miles,    with  the 
privilege    of   taking    in    three   more   persons.     Having   gone  20 
miles,  they  took  in  A  ;  at   Concord   they   took   in   B  ;   and  when 
within  30  miles  of  Boston,   they   took   in   C.     How   much    shall 
each  pay?     Ans.     First  man,  $7.609|£f;  second,   $7.609  W  ;  A» 
$5.873TV¥;  B,  $2,864^  ;  and  C,  $1.041TV 

11.  Three  men  purchased   6750  sheep.     The  number  of  A's 
sheep  is  to  the  number  of  B's  sheep  as  f  is  to  3^,  and  4  times  the 
number  of  C's  sheep  is  to  the  number  of  A's  sheep  as  -J   is    to  -J-. 
Find  the  number  of  sheep  each  had.  C  A's= 

Ans.  ]  B's  = 
(  C's  = 

12.  If  $500  gain  $10  in  4  months,  what  is  the   rate   per  cent? 

Ans.  6%. 

13.  If  12  men  can  do  as  much  work  as  25  women,  and   5  wo- 
men do  as  much  as  6  boys  ;  how  many  men  would   it  take  to  do 
the  work  of  75  boys?  Ans.  30  men. 

14.  If  5  experienced  compositors  in    16   days,    11   hours  each, 
can  compose  25  sheets  of  24  pages  in  each   sheet,  44   lines   on  a 
page,  8  words  in  a  line,  and  5  letters  to  a  word  ;   how   many  in- 
experienced compositors  in  12  days,  10  hours  each,   will   it  take 
to  compose  a  volume  (to  be  printed  with  the  same  kind  of  type), 
consisting  of  36   sheets,   16   pages   to   a   sheet,   112   lines   to  the 
page,  5  words  to  a  line,  and  8    letters   to  a   word,   provided  that 
while  composing  an  inexperienced  compositor  can  do   only  ^  as 
much  as  an  experienced  compositor,  and  that   the   latter  work  is 
only  f  as  hard  as  the  former?  Ans.  16. 

15.  If  A  can  do  f-  as  much  in  a  day  as  B,  B  can  do  f  as  much 
as  C,  and  C  can  do  |-  as  much  as  D,  and  D  can  do  -|  as   much  as 
E,  and  E  can  do  f  as  much  as  F;  in  what  time  can  F  do  as  much 
work  as  A  can  do  in  28  days?  Ans.  8. 

16.  A  starts  on  a  journey,  and  travels  27  miles  a  day;   7  days 
after,  B  starts,  and  travels  the  same,  road,  36  miles  a  day;  in  how 
many  days  will  B  overtake  A?  Ans.  21  days. 


146  FINKEL'S   SOLUTION   BOOK. 

17.  A  wheel  has  45  cogs  ;  a  smaller  wheel  working   in   it,  36 
cogs ;   in   how   many   revolutions   of  the   larger   wheel   will  the 
smaller  gain  10  revolutions?  Ans.  40. 

18.  If  the  velocity  of  sound  be  1142  feet  per   second,   and  the 
number  of  pulsations  in  a  person  70  per  minute,  what  is  the  dis- 
tance of  a  cloud,  if  30  pulsations  are  counted  between  the  time  of 
seeing  a  flash  of  lightning  and  hearing  the  thunder? 

Ans.  5|  mi.  108  yd.  If  ft 

19.  If  William's  services  are  worth  $15f-  a   month,   when  he 
labors  9  hours  a  day,  what  ought   he   to   receive   for   if  months, 
when  he  labors  12  hours  a  day?  Ans   $91.91^. 

20.  If  300  cats  kill  300  rats  in   300  minutes,   how   many  cats 
will  kill  100  rats  in  100  minutes?  Ans.  300  cats. 


CHAPTER    XVII. 

ANALYSIS. 

1.  Analysis,  in  mathematics,  is  the  process  of  solving 
problems  by  tracing  the  relation  of  the  parts. 

I.     What  will  7  lb.  of  sugar  cost  at  5  cents  a  pound? 
Analysis  for  primary  classes. 

If  one  ponnd  of  sugar  costs  5  cents,  7  pounds  will  cost  7  times 
5  cents,  which  are  35  cents. 

I.  If  6  lead  pencils  cost  30  cents,  what  will  one  lead  pencil 
cost? 

Analysis:  If  6  lead  pencils  cost  30  cents,  one  lead  pencil  will 
cost  as  many  cents  as  6  is  contained  into  30  cents  which  are  5 
cents. 

I.     If  8  oranges  cost  48  cents,  what  will  5  oranges  cost? 

Analysis:  If  8  oranges  cost  48  cents,  one  orange  will  cost  as 
many  cents  as  8  is  contained  into  48  cents  which  are  6  cents;  if 
one  orange  costs  6  cents  5  oranges  will  cost  5  times  6  cents,  which 
are  30  cents. 

I.  If  a  boy  had  7  apples  and  ate  2  of  them,  how  many  had  he 
left? 

Analysis:  If  a  boy  had  7  apples  and  ate  2  of  them,  he  had 
left  the  difference  between  7  apples  and  2  apples  which  are  5 
apples. 

I.  If  John  had  12  cents  and  found  5  cents,  how  many  cents 
did  he  then  have? 

Analysis:  If  John  had  12  cents  and  found  5  cents,  he  then, 
had  the  sum  of  12  cents  and  5  cents  which  are  17  cents. 


ANALYSIS.  147 

Note. — If  teachers  in  the  Primary  Departments  would  see  that  their 
pupils  gave  the  correct  analysis  to  such  problems,  their  pupils  would  often 
be  better  prepared  for  the  higher  grades.  After  they  are  thoroughly  ac- 
quainted with  the  analysis  of  such  questions  they  may  be  taught  to  write 
out  neat,  accurate  solutions  with  far  less  trouble  than  if  allowed  to  give 
careless  analysis  to  problems  in  the  lower  grades. 

I.     If  4  balls  cost  36  cents,  how  many  balls  can  be  bought  for 
81  cents? 

Analysis:  If  4  balls  cost  36  cents,  one  ball  will  cost  as  many 
cents  as  4  is  contained  into  36  cents  which  are  9  cents;  if  one  ball 
costs  9  cents  for  81  cents  there  can  be  bought  as  many  balls  as  9 
is  contained  into  81  which  are  9  balls. 

Written  solution. 

(  1.  36  cents=cost  of  4  balls. 
II.  \  2.  9  cents=36  cents-^4=cost  of  1  ball. 
(  3.  81  cents=cost  of  81-=-9,  or  9  balls. 

III.     /.  If  4  balls  cost  36  cents,  for  81  cents  there  can  be  bought 
9  balls. 

I.     What  number  divided  by  -|  will  give  10  for  a  quotient? 


II. 


1.  |r=the  number. 

2-  *-H=*  X4=f=quotient 


3.  10=quotient. 

4.  /,  1=10, 

5.  £=£of  10=2,  and 

6.  =3  times  2=6=the  number. 


III.      .'•'.  6=the  number  required. 

I.     $24  is  f  of  the  cost  of  a  barrel  of  wine;  what   did   it  cost? 

II.  |=cost  of  the  wine  per  barrel. 
2.  \  of  cost=$24, 
3.  \  of  cost=£  of  $24=$8, 
4.  |  of  cost=5  times  $8=$40, 

(II.     .%  $40=cost  of  wine. 

t.     What  number  is  that  from  which,  if  you   take  \  of  itself, 
the  remainder  will  be  16  ? 

1.  ^=the  number. 

2.  If — f=4=remainder  after  taking  away  ^. 

3.  16=remainder. 
.-.$=16, 

5.  f=i  of  16=4,  and 

6.  ^=7  times  4=28=the  number 

III.      .'.  28=the  required  number. 


148  FINKEL'S   SOLUTION  BOOK. 

I.  A  boat  is  worth  $900;  a  merchant  owns  £  of  it,  and  sells  % 
of  his  share  ;  what  part  has  he  left,  and  what  is  it 
worth  ? 

1.  ^=part  the  merchant  owned. 

2.  |  of  !=^=part  he  sold. 

i  ,          3.  .-.  f-AHri—  &=M=T\=Part  he  had  left. 

-1     $900=value  of  Tf  ,  or  the  whole  ship. 
R     ,  *.  $75=TV  of  $900=value  of  ^  of  the  ship. 

1  3.  $375=5  times  $75=value  of  ^  of  the  ship,  or  part 
he  had  left. 


$375=value  of  it. 

I.  A  and  B  were  playing  cards.  B  lost  $14,  which  was  y7^ 
times  f  as  much  as  A  then  had  ;  and  when  they  com- 
menced, |  of  A's  money  equaled  fy  of  B's.  How  much 
had  each  when  they  began  to  play? 

(  1.  )         |  of  A's  money=4  of  B's. 

(2.  )         J  of  A's  money=|  of  f  =-£%  of  B's. 


II. 


=B's  money  when  they  began   to   play.     Then 


(3.)  f  of  A's  money=8  times  A=M  of  B's- 

(4.) 

(5.)  ^|.=A's  money  when  they  began. 

1.  Tf=A's  money  after  winning  $14  from  B. 

2.  $14=what  B  lost. 

3.  T7V  times  f=T7^^part  A's  money  is  of  $14. 

5.  ^=4  of  $14=$2,  and  [$14  from  B. 

6.  If— 15  times  $2=$30=A's  money  after  winning 
(7.)  .'.  $30— $14=$  16= A's  money  at  first. 


.) 

.) 


(8.)         .-.  if=$16,  from  (5), 

^(10.)         H=35  times  $l=$35=B's  money  at  first. 

$16=A's  money  at  first,  and 
$35=B's  money  at  first. 

(Stod.  Int.  A., p.  lll.prob.  30.) 

I.  A  drover  being  asked  how  many  sheep  he  had,  said,  if  to 
•J  of  my  flock  you  add  the  number  9^-,  the  sum  will  be 
99^-;  how  many  sheep  had  he? 

'1.  4|=the  number  of  sheep. 

2.  i+9-^=-J  of  the  number-|-9^. 

3.  99£=£  of  the  number+9f 

5.  4=99^—9-^=90,  and 

6.  f=3  times  90=270=number  of  sheep. 

III.     .'.  He  had  270  sheep. 


ANALYSIS. 


149 


I. 


II. 


III. 


I. 


II. 


III. 


I. 


Heman  has  6  books  more  than  Handford,  and  both  have 
26;  how  many  have  each? 

1.  |=number  Handford  has.    Then 

2.  |+6=Heman's  number. 

3.  f +f +6=|+6=number  both  have. 

4.  26=number  both  have. 

6.  .-.  f+6=26  or 

5.  4^26—6=20. 

7.  f=iof20=6,  and 

8.  f=2  times  5=10=Handford's  number. 

9.  f+6=16=Heman's  number. 

Handford  had  10  books,  and 

Heman  had  16  books.     (Stod.  Int.  A., p.  116,prob.  2.) 

A  man  and  his  wife  can  drink  a  keg  of  wine  in  6  days, 
and  the  man  alone  in  10  days  ;  how  many  days  will  it 
last  the  woman? 

1.  6  days=time  it  takes  both  to  drink  it. 

2.  ^=part  they  drink  in  one  day. 

3.  10  days=time  it  takes  the  man  to  drink  it. 

4.  T1^=part  he  drinks  in  one  day.  [day. 

5.  .'.  ^ — 'fu=='f^ — A==T"5==:Par*  *ne  woman  drinks  in  one 

6.  -j~!=what  the  woman  drinks  in  ±-^-±-^=1.5  days. 


II. 


III. 


.•.  It  will  take  the  woman  15  days. 

(/?.  Alg.  /.,/.  112,prob.  59.) 

A  man  was  hired  for  80  days,  on  this  condition:  that  for 
every  day  he  worked  he  should  receive  60  cents,  and 
for  every  day  he  was  idle  he  should  forfeit  40  cents.  At 
the  expiration  of  the  time,  he  received  $40.  How  many 
days  did  he  work? 

'1.  $.60=what  he  receives  a  day. 

2.  $48=80  X$.  60=  what  he  would   have   icceived   had   he 

worked  the  whole  time. 

3.  $40=what  he  received. 

4.  ...  $48_$40=$8=what  he  lost  by  his  idleness. 

5.  $1=$.60,    his    wages,+$.40,  what    he   had    to  forfeit,= 

what  he  lost  a  day. 

6.  /.  $8=what  he  lost  in  8-f-l,  or  8  days. 

.  80  clays  —  8  days=72  days,  the  time  he  worked. 

/.  He  worked  72  days. 

A  ship-mast  51  feet  high,  was  broken  off  in  a  storm,  and 
|  of  the  length  broken  off,  equaled  f  of  the  length  re- 
maining; how  much  was  broken  off,  and  how  much  re- 
mained? 


150  FINKEL'S   SOLUTION   BOOK. 

1.  f  of  length  broken  off=f  of  length  remaining, 

2.  -J  of  length  broken  off=£  of  f=|  of  length  remaining, 

3.  f  of  length  broken  off=3  times  f=f  of  length  remain- 

ing. 

4.  f— length  remaining. 


II. 


II. 


j 


5.  f— length  broken  off. 

6.  f-f-f=y=whole  length. 

7.  51  feet=whole  length. 

8.  /.  V=51  feet, 

9.  4=TV  of  51  feet=3  feet,  and 

10.  f=8  times  3  feet=24  feet,  length  remaining. 

11.  |=9  times  3  feet=27  feet,  length  broken  off. 

24  feet— length  remaining,  and 
27  feet=le*igth  broken  off. 

I.     A  boy  being   asked   his   age,  said,  "4  times   my  age  is  24 
years  more  than  2  times  my  age;"  how  old  was  he? 

1.  !=his  age. 

2.  4xf=f=4  times  his  age. 

3.  2Xf=|=2  times  his  age. 

4.  .-.  |=|._|_24  years  or 

5.  | — 4^4^24  years. 

6.  i=|-  of  24  years=6  years,  and 

L7.  f=2  times  6  years=12  years,  his  age. 

III.     /.  He  is  12  years  old.          (Stod.  Int.  A., p.  116,prob.  16.) 


If  10  men  or  18  boys  can  dig  1  acre  in  11  days,  find  the 
number  of  boys  whose  assistance  will  enable  5  men  to 
dig  6  acres  in  6  days. 

•  1.  1  A.=what  10  men  dig  in  11  days. 

2.  YV  A.=what  1  man  digs  in  11  days. 

3.  YTO  A.=YT  °f  TTT  A.— what  1  man  digs  in  1  day. 

4.  Y2"  A.=YTTF  A.=5  times  YYTF  A.=what  5  men  dig  in  1 

day.  [days. 

5.  fT  ^.=2%  A.=6   times  ^  A.=what  5  men   dig  in  6 

6.  .'.  6  A. — T3T  A.=5T8i  A.=what  is  to  be  dug  by  the  boys 

in  6  days. 

7.  1  A.=what  18  boys  dig  in  11  days. 

8.  Y*g-  A.=what  1  boy  digs  in  11  days. 

9.  Yi~g"  ^•==TT  °f  Tff  A.=what  1  boy  digs  in  1  day. 

10.  -g^  A.=Tf'j  A.=6  times  T^-g-  A.=what  1  boy  digs  in  6 

days. 
-11.  5-fa  A.=what  5T8Y-:~g^,  or  189,  boys  dig  in  6  days. 

III.     .'.  It  will  take  198  boys. 

(R.  3d  p.,  O.  E.,p.  318,prob.  66.) 


ANALYSIS.  151 

A  man  after  doing  |  of  a  piece  of  work  in  30  days,  calls 
an  assistant;  both  together  complete  it  in  6  days.  In 
what  time  could  the  assistant  complete  it  alone? 

1.  -|==part  the  man  does  in  30  days. 

2.  -ffa—^  of  f—  part  he  does  in  1  day. 

3.  -f=f  —  |=part  he  and  the  assistant  do  in  6  days. 

4.  T15=-g-  of  -|=part  he  and  the  assistant  do  in  1  day. 

&•  •'•  T*5  —  5Tff=rA  —  rTr=Ti7r==Part  the  assistant  does  in  1 

day. 
6.  ff$=part  the  assistant  does  in  -J-J^-i-^^lf  days. 


II. 


III.     /.  It  will  take  the  assistant  21f  days. 

(R.  3d  /.,  O.  E.,p.  318,prob.  71.) 

Explanation^.-  —  Since  the  man  does  f  of  the  work  before  he  called  on  the 
assistant,  there  remains  f  —  1=§,  which  he  and  the  assistant  do  in  6  daj  s. 
Hence  they  do  £  of  §,  or  ^  of  the  work  in  one  day.  If  the  man  and  his^ 
assistant  do  ^  of  the  work  in  1  day  and  the  man  does  -£$  of  the  work  in  1 
day,  the  assistant  does  the  difference  between^  and  ^  which  is  r£0  of  the 
work  in  1  day.  Hence  it  will  take  £$8-7-r£o>  or21f  days,  to  do  the  work. 

L  A  person  being  asked  the  time  of  day,  replied  that  it  was- 
past  noon,  and  that  f  of  the  time  past  noon  was  equal 
to  f  of  the  time  to  midnight.  What  was  the  time  of 
day? 

1.  |  of  the  time  past  noon=f  of  the  time  to  midnight. 

2.  i  of  the  time  past  noon=4  of  -f=£  of  the  time   to  mid- 

night. [midnight. 

3.  |>  or  the  time  past  noon,=4  times  -$=-f  of  the   time  to 

4.  f=time  to  midnight.     Then 


II. 


5.  -|=time  past  noon. 


6.  f-j—  |—  ^=time  from  noon  to  midnight. 

7.  12  hours=time  from  noon  to  midnight. 

8.  .*.  |=12  hours, 

9.  -j=7f  of  12  hours=l^  hours,  and  [past  noon. 
10.  ^=4    times   1^-   hours=5^-  hours=5   hr.   20  min.,  time 


III.     .-.  It  is  20  min.  past  5  o'clock,  P.  M. 

(Mztne's  Prac.  A., p.  S60,  prob.  47.) 

Note. — From  3,  we  have  the  statement  that  the  time  past  noon  is  £  of  the 
time  to  midnight.  Hence,  if  f  is  the  time  to  midnight,  £  is  the  time  past 
noon  or  if  {§  is  the  time  to  midnight,  ^  is  the  time  past  noon. 

I.  A  person  being  asked  the  time  of  day,  said  that  ^  of  the 
time  past  noon  equals  the  time  to  midnight.  What  is 
the  time  of  day? 


152 


FINKEL'S   SOLUTION  BOOK. 


1.  -5-=time  past  noon.     Then 

2.  ^=time  to  midnight. 

3.  5— [-7.=  1T2=time  from  noon  to  midnight. 
Ut<J  4.  12  hours=time  from  noon  to  midnight. 

5.  .'.  J72=12  hours. 

6.  Jf—Tz  of  12  hours=l  hour,  and 

7.  -J=7  times  1  hour=7  hours=time*past  noon. 

III.     .-.  It  is  7  o'clock  P.  M. 

I.  A  man  being  asked  the  hour  of  day,  replied  that  £  of  the 
time  past  3  o'clock  equaled  \  of  the  time  to  midnight; 
what  was  the  hour? 

1.  £  of  the  time  past  3  o'clock=-J  of  the  time  to  midnight. 

2.  |,  or  the  time  past  3  o'clock,=4  times  -J=f  of  the  time 

to  midnight. 

3.  |=:time  to  midnight. 

4.  |-=time  past  3  o'clock. 

H.-I    5.  f+f =£=time  from  3  o'clock  to  midnight. 

6.  9  hours=time  from  3  o'clock  to  midnight. 

7.  .-.  |=9  hours. 

8.  \-=\  of  9  hours=l^  hours,  and 

9.  |=4  times  1£  hours=6  hours=time  past  3  o'clock. 
.10.  f+3  hours=9  hours,  time  past  noon. 

III.     .-.  It  is  9,  o'clock,  P.  M. 

(Brooks1  Int.  A.,  p.  156,prob.  17.) 

I.  A  person  being  asked  the  hour  of  day,  replied,  f  of  the 
time  past  noon  equals  |-  of  time  from  now  to  midnight 
+2f  hours;  what  was  the  time? 

1.  f  of  time  past  noon=J  of  time  to   midnight-)-2f  hours. 

2.  J  of  time  past  noon=-J  of  (f+2|  hours )=^  of  time  to 

midnight-]- 1£  hours.  [to  midnight-f-4  hours. 

3.  f,  or  time  past  noon,=3  times  (^-[-1-j- hours )=-J  of  time 

4.  f=time  to  midnight. 

I  J    5.  £+4  hours=time  past  noon.  [night. 

6.  f-HH~4  hours=:|-(-4  hours=time   from  noon   to  mid- 

7.  12  hours=time  from  noon  to  midnight. 

8.  .'.  f-f-4  hours=12  hours. 

9.  £==12  hours — 4  hours=8  hours, 

10.  |-=i  of  8  hours=2  hours,  and  * 

11.  |~|-4  hours=6  hours=time  past  noon. 

III.     .-.  It  is  6  o'clock,  P.  M. 

(Stod.  Int.  A., p.  128,  Prob.  29.) 

I.  A  father  gave  to  each  of  his  sons  $5  and  had  $30  remain- 
ing; had  he  given  them  $8  each,  it  would  have  taken  all 
his  money;  required  the  number  of  sons, 


ANALYSIS. 


153 


'1.  $8=amount  each  received  by  the  second  condition. 
2.  $5=amount  each  received  by  the  first  condition. 
II.  •{  3.  $3=$8 — $5=  excess  of  second   condition  over  first,  on 
each  son.  [10  sons. 

,4.  /.  $30=excess  of  second  condition  over  first,  on  30-i-3,  or 

I.     .•.  There  were  10  sons. 

I.  If  50  lb.  of  sea  water  contain  2  Ib.  of  salt,  how  much  fresh 
water  must  be  added  to  the  50  lb.  so  that  10  lb.  of  the 
new  mixture  may  contain  -J  lb.  of  salt. 

1.  -J,  lb.  of  salt=what  10  lb.  of  the  new  mixture  contains. 

2.  f ,  or  1,  lb.  of  salt=what  3  times  10  lb.,  or  30  lb.,  of  the 
,,  i  new  mixture  contain.  [mixture  contain. 

X  3.  2  lb.  of  salt=what  2  times  30  lb.,  or  60  lb.,   of  the  new 
4.  .-.  60  lb.— 50  lb.=10  lb.=quantity  of  fresh   water  that 
must  be  added. 


III. 


I. 


II. 


.-.  10  lb.  of  fresh  water  must  be  added  that  10  lb.  of  the 
new  mixture  may  contain  -j-  lb.  of  salt. 

A  farmer  had  his  sheep  in  three  fields,  f  of  the  number 
in  the  first  field  equals  f  of  the  number  in  the  second 
field,  and  f-  of  the  number  in  the  second  field  equals  £ 
of  the  number  in  the  third  field.  If  the  entire  num- 
ber was  434,  how  many  were  in  each  field? 

•§•  of  number  in  first  field=f  of  number  in  second 
field.  [second  field. 

-J-  of  number  in  first  field=-J-  of  f=|  of  number  in 
-| ,  or  number  in  first  field,=3  times  f=f  of  number 

in  second  field. 

|  of  number  in  second  field=|  of  number  in  third 
field.  [in  third  field. 

•J  of  number  in  second  field=-J-  of  f =f  of  numbet 
f,  or  number  in  second  field,=3  times  f=f  of  num- 
ber in  third  field. 
(3.)         f=number  in  third  field.     Then 
(4.)         -|=number  in  second  field,  and 
(5.)        £4=t  of  number  in  second  field=number  in  first 
field  in  terms  of  number  in  third  field. 

three  fields. 

434=number  in  the  three  fields. 

...  2^=434, 

_T  of  434=2,  and  [field. 

[=64  times  2=128=number  of  sheep  in  third 
5=72  times  2=144=number  of  sheep  in  second 
field.  [field. 

1^=81   times   2=162=number   of  sheep   in   first 


154 


FINKEL'S  SOLUTION  BOOK. 


III. 


II. 


I. 


II. 


C  162=number  of  sheep  in  first  field, 

<  144=number  of  sheep  in  second  field,  and 

(  128=number  of  sheep  in  third  field. 

(Milne's  Prac.  A., p.  362,  prob.  68.) 


pupils  there   are   32  girls ;  how 
that   there  may  be  5  boys   to  4 


In  a  certain  school  of  80 
many  boys  must  leave 
girls? 

1.  80=whole  number  of  pupils. 

2.  32— number  of  girls. 

3.  go — 32=48=number  of  boys. 

4.  ^=number  of  girls.     Then,  since  the  number  of  boys  are 

to  be  to  the  number  of  girls  as  5  :  4, 

5.  J=number  of  boys.     But 

6.  |=32. 

7.  £=£  of  32=8,  and 

8.  4=5  times  8=40=number  of  boys. 

9.  .'.  48 — 40=8=number  that  must  leave  that  there  may  be 

5  boys  to  4  girls. 

III.     .'.  8  bo^s  must  leave  that  there  may  be  5  boys  to  4  girls. 


How  far  may  a  person  ride  in  a  coach,  going  at  the  rate 
of  9  miles  per  hour,  provided  he  is  gone  only  10  hours, 
and  walks  back  at  the  rate  of  6  miles  per  hour? 

1.  9  mi.=distance  he  can  ride  in  1  hour. 

2.  1  mi.=distance  he  can  ride  in  -J-  hour. 

3.  6  mi.=distance  he  can  'walk  in  1  hour. 

4.  1  mi.=distance  he  can  'walk  in  \  hour. 

hr.-)-^  hr.=T5g-  hr.=time  it  takes  him   to  ride   1  mi. 
and  walk  back.  [and  walk  back. 

10  hours=time  it  takes  him  to  ride  10-7-T5^,  or  36,  mi. 


5. 


6. 


III.     .-.  He  can  ride  36  miles. 

I.  A  hound  ran  60  rods  before  he  caught  the  fox,  and  f  of 
the  distance  the  fox  ran  before  he  was  caught,  equaled 
the  distance  he  was  ahead  when  they  started.  How  far 
did  the  fox  run,  and  how  far  in  advance  of  the  hound 
was  he  when  the  chase  commenced? 

rl.  f=distance  the  fox  ran  before  he  was  caught.     Then 

2.  -|=distance  he  was  ahead. 

3.  f +f=|=distance  the  hound  ran  to  catch  the  fox. 

4.  60  rods=distance  the  hound  ran  to  catch  the  fox. 

5.  .-.  f=60  rods, 

6.  -J=^  of  60  rods=12  rods,  and  [ahead. 

7.  f=2    times    12    rods=24    rods=distance   the   fox   was 

8.  f=3  times  12  rods=36  rods=distance  the   fox  ran  be- 

fore he  was  caught. 


ANALYSIS. 


155 


(  24  rods=distance  the  fox  was  ahead,  and 

(  36  rods=distance  he  ran  before  he  was  caught. 

If  J  of  12  be  3  ,  what  will  ±  of  40  be  ? 

1.  •§•  of  12=4. 

2.  i  of  40=10.     By  supposition 

3.  4=3.     Then 

4.  1=4  of  3=|,  and 

5.  10=10  times  |=7|. 

.-.  £  of  40=7  -J,  on  the  supposition  that  £  of  12  is  3. 

Eight  men  hire  a  coach;  by  getting  6  more  passengers, 
the  expenses  of  each  were  diminished  $1|;  what  do  they 
pay  for  the  coach? 

1.  f—  amount  paid  for  the  coach.  [been  only  8  men. 

2.  -£=amount    1   man   would   have   had   to  pay,  had  there 

3.  T1?=amount  1  man  paid  since  there  were  8  men-f-6  men, 

or  14  men. 

4.  .'•  i—  TT=T6—  Tnr=irs=what  each  saved. 

5.  $lf=what  each  saved. 

6.  .-. 


7.     =    of  $i=$7>  and 


II. 


18.  ff=56  times  $-J2^==$32|=amount  paid  for  the  coach. 

HI.     .%  $32f=amount  paid  for  the  coach. 

(R.  H.  A., p.  Jt.OS.prob.  46.) 

Second  solution. 

'1.  $lf=amount  saved  by  each  man.  [the  six  meu. 

2.  $14=8X$lf=amount  saved  by  the  8  men  and  paid  by 

3.  .*.  $2£=J  of  $14=amount  paid  by  each  of  the  14  men. 
,4.  .-.  $32f— 14  times  $2£=amount  they  paid  for  the  coach; 

.-.  They  paid  $32f  for  the  coach. 

For  every  10  sheep  I  keep  I  plow  an  acre  of  land,  and 
allow  one  acre  of  pasture  for  every  4  sheep;  how  many 
sheep  can  I  keep  on  161  acres? 

1.  1  A.=what  I  plow  for  every  10  sheep  I  keep. 

2.  T^A.=what  I  plow  for  each  sheep  I  keep. 

3.  1  A.=what  I  allow  for  pasture  for  every  4  sheep  I  keep. 

4.  ^A.=what  I  allow  for  pasture  for  each  sheep  I  keep. 

5.  .'.  T1irA.-|-JA.=^FA.=land  required  for  every  sheep. 

6.  c-.  161A.=land  required  for  161-r-^r,  or  460  sheep. 

.-.  I  can  keep  460  sheep  on  161  acres. 

(./?.  Alg.  I.) p.  112,prob.  64') 
Complete  analysis. 

If  for  every  10  sheep  I  plow  1  acre,  for  1  sheep  I  plow  -^  of 
&n  acre  ;  and'  if  for  every  4  sheep  I  pasture  1  acre,  for  1  sheep,  I 


III. 
I. 


II. 


III. 


156  FINKEL'S  SOLUTION  BOOK. 

pasture  ^  of  an  acre  ;  hence  1  sheep  requires  ^A.-j-^A.,  or  ^.n.., 
and  on  161  A.  I  could  keep  as  many  sheep  as  /^A.  is  contained  in, 
161  A.,  which  are  460  sheep. 

I.  A  man  was  engaged  for  one  year  at  $80  and  a  suit  of 
clothes;  he  served  7  months,  and  received  for  his  wages 
the  clothes  and  $35;  what  was  the  value  of  the  clothes? 

1.  i|=value  of  the  suit  of  clothes. 

2.  ff4-$80=wages  for  1  year  or  12  months. 

3.  ^^iej^^  of  (ff+$80)=:wages  for  1  month. 

4.  ^-|-$46|=7  times  (TVf$6|-)== wages  for  7  months. 
!!.<!  5.  ff +$ 3 5= wages  for  7  months. 

6.  /.  -if +$35=T^+$46|-. 

8.  T%=-%  of  $11^=::=$2^-,  and 

9.  ff=12  times  $2£=$28=value  of  suit  of  clothes. 


III.     /.  The  suit  of  clothes  is  worth 


A  lady  has  two  silver  cups,  and  only  one  cover.  The 
first  cup  weighs  8  ounces.  The  first  cup  and  cover 
weighs  3  times  as  much  as  the  second  cup;  and  the  sec- 
ond cup  and  cover  4  times  as  much  as  the  first  cup. 
What  is  the  weight  of  the  second  cup  and  the  cover? 

1.  3  times  weight  of  second  cup=weight  of  cover-)-  weight 

of  first  cup,  or  8  oz.  [2f  oz. 

2.  1  times  weight  of  second  cup=£  of  weight  of  cover-J- 

3.  |=weight  of  cover.     Then 

4.  -J+2f  oz.—  weight  of  second  cup.  [cover. 

5.  I+J+2J  oz.=£-|-2f  oz.  =  weight   of  second  cup  and 
!!.<{   6.  32  oz.—  4  times  8  oz.=weight  of  second  cup  and  cover, 

by  the  conditions  of  the  problem. 

7.  .'.  i+2f  oz.=32  oz. 

8.  |=32  oz.—  2f  oz.=29£  oz. 

9.  |=i  of  29£  oz.=7£  oz. 

10.  f=3  times  7^-  oz.=22  oz.=weight  of  cover.  [cup. 

11.  l+2f  oz.=7l  oz.+2|  oz.=  10  oz.  =  weight  of   second 

.    (  22  oz.=weight  of  cover,  and 
*  "  (  10  oz.=weight  of  second  cup. 


I.  A  steamboat  that  can  run  15  mi.  per  hr.  with  the  current 
and  10  mi.  per  hr.  against  it,  requires  25  hr.  to  go  from 
Cincinnati  to  Louisville  and  return  ;  what  is  the  dis- 
tance between  the  cities? 


ANALYSIS.  157 

1.  15  mi.=distance   the   boat   can    travel    down    stream   in 

1   hour.  [hour. 

2.  1  mi.=distance  the  boat  can  travel   down   stream   in  -^ 

3.  10  mi.— distance  the  boat  can  travel  up   stream   in   1  hr. 
II. <  4.   1  mi.=distance  the  boat  can  travel  up  stream   in   -fa  hr. 

5.  /.  y^-  hr.-l-y1^  hr.=^r  hr.=time  required   for   the   boat  to 

travel  1  mi.  down  and  return. 

6.  .'.  25  br.=time  required  for  the  boat  to  travel  25-r-J-,  or 

150,  milss  down  and  return. 

III.      /.  The  distance  between  the  two  places  is  150  miles. 

I.  A,  B,  and  C  dine  on  8  loaves  of  bread  ;  A  furnishes  5 
loaves  ;  B,  3  loaves;  C  pays  the  others  8d.  for  his  share; 
how  must  A  and  B  divide  the  money  ? 

1.  8  loaves=what  they  all  eat. 

2.  2f  loaves— what  each  eats. 

3.  .'.5  loaves — 2f    loaves=2-j-   loaves=what    A    furnished 

towards  C's  dinner. 

4.  .-.   3    loaves — 2f   loaves=4   loaf=what   B   furnished  to- 

wards C's  dinner. 

5.  .'.  — |=-|=A's  share,  and 

6.  /.  ^-— -J— B's  share. 

7.  |  of  8d.=7d.=what  A  should  receive,  and 
.8.  ^  of  8d.=ld.=what  B  should  receive. 

Ill       •     \  ^  should  receive  7d.,  and 

'    I  B  should  receive  Id.  (R.  H.  A.,  p.  403,  prob.  42.) 

I.  A  and  B  dig  a  ditch  100  rods  long  for  $100;  how  many 
rods  does  each  dig,  if  they  each  receive  $50,  and  A  digs  at  $.75 
per  rod,  and  B  at  $1.25? 

There  has  been  a  vast  amount  of  quibbling  about  this  problem; 
but  a  few  moments  consideration  should  suffice  to  settle  all  dis- 
pute, and  pronounce  upon  it  the  sentence  of  absurdity. 

We  have  given,  the  whole  amount  each  received  and  the 
amount  each  received  per  rod.  Hence,  if  we  divide  the  whole 
amount  each  received  by  the  cost  per  rod,  it  must  give  the  num- 
ber of  rods  he  digs.  But  by  doing  this  we  receive  50-—.75,  or  66f 
rods,  what  A  digs  and  50-^-1.25,  or  40  rods,  what  B  digs,  or 
106f  rods  which  is  the  length  of  the  ditch,  and  not  100  rods  as 
stated  in  the  problem.  The  length  of  the  ditch  is  a  function  of 
the  cost  per  rod  and  the  whole  cost,  and  when  they  are  given 
the  length  of  the  ditch  is  determined.  We  might  propose  a 
problem  just  as  absurd  by  requiring  the  circumference  of  a  circle 
whose  area  is  1  acre,  and  diameter  20  rods.  Since  the  area 
and  circumference  are  functions  of  the  diameter,  when  either 


158  FINKEL'S   SOLUTION  BOOK. 

of  these  are  given,  the  other   is  determined    and    should   not  be 
limited  to  an  inaccurate  statement. 

If,  in  the  original  problem,  A's  price  per  rod  increases  at  a 
constant  ratio  so  that  when  the  ditch  is  completed  he  is  receiving 
$1  per  rod,  and  B's  price  constantly  decreases  until  when  the 
ditch  is  completed  he  is  receiving  $1  per  rod,  then  the  problem 
is  solvable,  and  the  result  is  50  rods  each. 

I.  A  is  30  years  old,  and  B  is  6  years  old  ;  in  how  many 
years  will  A  be  only  4  times  as  old  as  B? 

!=B's  age  at  the  required  time.     Then 
•|=A's  age  at  the  required  time. 

3.  | — f=J=difference  of  their  ages. 

4.  30  years — 6  years=24  years=difFerence  of  their  ages. 
II.<5.  .*.  f— 24  years. 

6.  -i=-jr  of  24  years=4  years.  [time. 

7.  1=2   times   4   years— 8   years,  B's   age   at   the  required 

8.  .-.  8  years — 6  years— 2  years=the  number  of  years  hence 

when  A  will  be  only  4  times  as  old  as  B. 

III.     .-.  In  2  years  A  will  be  only  4  times  as  old  as  B. 

I.  Jacob  is  twice  as  old  as  his  son  who  is  20  years  of  age  ; 
how  long  since  Jacob  was  5  times  as  old  as  his  son? 

1.  20  years=son's  age  at  present.     Then 

2.  40  years=Jacob's  age  at  present. 

3.  |-:=son's  age  at  required  time.     Then 

4.  y)=Jacob's  age  at  required  time. 

5.  .'.  l-£ — f — f=difference  of  their  ages. 

II.<!    6.  40  years — 20  years=20  years=difference  of  their  ages. 

7.  ...  f=20  years, 

8.  -J=^  of  20  years=2^  years,  and  [time. 

9.  |=2  times  2-J  years=5  years,  son's  age  at  the  required 
10.  /.  20  years — 5  years— 15  years=time  since  Jacob  was  5 

times  as  old  as  his  son. 
HI.     .'.  15  years  ago  Jacob  was  5  times  as  old  as  his  son. 

Remarks. — Observe  that  the  difference  between  any  two  persons'  ages  is 
constant,  that  is,  if  the  difference  between  A's  and  B's  ages  is  7  years  now, 
it  will  be  the  same  in  any  number  of  years  from  now;  for,  as  a  year  is  add- 
ed to  one's  age,  it  is  likewise  added  to  the  other's  age.  But  the  ratio  of 
their  ages  is  constantly  changing  as  time  goes  on.  If  A  is  3  years  old  and 
B  5  years  old,  A  is  now  £  as  old  as  B;  but  in  1  year,  A's  age  will  be  4  years 
and  B's  6  years;  A  is  then  f  as  old  as  B.  In  7  years,  A  will  be  10  years  old 
and  B  12;  A  will  then  be  {§,  or  f ,  as  old  as  B,  and  so  on.  The  ratio  of  any 
two  persons'  ages  approaches  unity  as  its  limit. 

I.  A  fox  is  50  leaps  ahead  of  a  hound,  and  takes  4  leaps  in 
the  same  time  that  the  hound  takes  3  ;  but  2  of  the 
hound's  leaps  equal  3  of  the  fox's  leaps.  How  many 
leaps  must  the  hound  take  to  catch  the  fox? 


ANALYSIS.  159 

1.  2  leaps  of  hound's=3  leaps  of  fox's. 

2.  1  leap  of  hound's=J  of  3  leaps=l£  leaps  of  the  fox's. 

3.  3  leaps  of  hound's=3  times  1£  leaps— 4^  leaps  of  fox's, 
j-  1 4.  .-.  4-J-  leaps — 4  leaps=£  leap=what  the  hound  gains  in 

taking  3  leaps.  [ing  6  leaps. 

5.  .'.  1  leap=2  times  •£•  leap=what  the  hound  gains  in  tak- 

6.  .'.   50  leaps=what  the  hound   gains    in    taking  50x6 

leaps,  or  300  leaps. 
III.     .-.  The  hound  must  take  300  leaps  to  catch  the  fox. 
Remark  — We  see  that  3  of  the  hound's  leaps  equals  4|  leaps  of  the  fox's, 
But  while  the  hound  takes  3  leaps,  the  fox  takes  4  leaps;  hence  the  hound 
gains  4£ — 4,  or  |,  leap  of  the  fox's.     But  he  has  50  leaps  of  the  fox's  to  gain, 
and  since  he  gains  \  leap  of  the  fox's  in  3  leaps,  he  must  take  300  leaps  to 
gain  50  leaps. 

I.     If  6  sheep  are  worth   2   cows,  and    10   cows   are   worth  5 
horses;  how  many  sheep  can  you  buy  for  3  horses? 

1.  Value  of  2  cows=value  of  6  sheep. 

2.  Value  of  1  cow=value  of  3  sheep. 

3.  Value  of  10  cows=value  of  30  sheep.     But  10  cows  are 


II. 


worth  5  horses, 


4.  /.  Value  of  5  horses=value  of  30  sheep. 

5.  Value  of  1  horse=value  of  6  sheep. 

6.  Value  of  3  horses=value  of  18  sheep. 
III.     /.  3  horses  are  worth  18  sheep. 

I.  A  teacher  agreed  to  teach  a  certain  time  upon  these  con- 
ditions :  if  he  had  20  scholars  he  was  to  receive  $25; 
but  if  he  had  30  scholars,  he  was  to  receive  but  $30. 
He  had  29  scholars.  Required  his  wages. 

1.  $25=his  rate  of  wages  for  20  pupils. 

2.  $1.25=2^  of  $25=his  rate  of  wages  for  1  pupil. 

3.  $30— his  rate  of  wages  for  30  pupils. 

4.  $l=^j-  of  $30— his  rate  of  wages  for  1  pupil. 

5.  /.  $1.25 — $1.00=$.25=reduction   per  pupil  by  the  ad- 


II. 


dition  of  10  pupils. 


6.  $.025— $.25-r-10=reduction  per  pupil  by  the  addition  of 

1  pupil. 

7.  $.225=9  times  $.025=reduction  per   pupil  by  the  addi- 

tion of  9  pupils. 

/.  $1.25— $.225=$1.025=his  rate  of  wage  per  pupil. 
9.  $29.725=29  times  $1.025=his  wages  for  29  pupils. 
III.     .-.  His  wages  were  $29.725. 

(Matt(*i?sArith.,p.  385,  f  rob.  200.) 

Note. — This  problem  is  really  indeterminate,  because  there  is  no  definite 
rate  of  increase  of  wages  given  for  each  additional  scholar.  We  might  say, 
since  the  wages  were  increased  $5  by  the  addition  of  10  scholars,  they  would 
be  increased  $.50  by  the  addition  of  one  scholar  and,  consequently,  $4.50  by 
the  addition  of  9  scholars.  Hence,  his  wages  should  be  $25+$4.50,  or 
$29.50.  By  assuming  different  relations  between  the  increase  of  wages  and 
additional  scholars,  other  results  may  be  obtained.  The  above  solution 
seems  to  be  the  most  satisfactory. 


160  FINKEL'S  SOLUTION  BOOK. 

I.  A  gold  and  silver  watch  were  bought  for  $160;  the  silver 
watch  cost  only  ^  as  much  as  the  gold  one  ;  how  mucb 
was  the  cost  of  each  ? 

1.  ^-=cost  of  the  gold  watch.     Then 

2.  ^=cost  of  the  silver  watch. 

3.  7.+i=8=cost  of  both. 

4.  $160=cost  of  both. 

5.  .'.  f=$160, 

6.  -J-=-J  of  $160='$20=cost  of  the  silver  watch,  and 

7.  7._7  times  $20=$140=cost  of  the  gold  watch. 

I  $20=cost  of  the  silver  watch,  and 
'    (  $140=cost  of  gold  the  watch. 

A  man  has  two  watches,  and  a  chain  worth  $20;  if  he  put 
the  chain  on  the  first  watch  it  will  be  worth  f  as  much 
as  the  second  watch,  but  if  he  put  the  chain  on  the  sec- 
ond watch  it  will  be  worth  2£  times  the  first  watch 
what  is  the  value  of  each  watch? 

1.  §  B.=J  f.+$20. 

2.  -k  s.=J-  of  (f  f.+$20)=J-  f.+$10. 

3.  f  s.= 3  times  (i  f.+$10)=£  f +$30.  [lem. 

4.  I  s.=1^  f. — $20,  by  the  second  condition   of  the  prob- 

II. 


5.  ...  11  f._$20=f  f.+$30,  whence 
y  f._ 4  f.=f 


6.  V  f-— f  f.=$30+$20,  or 

7.  f  f.— $50. 

8.  i  f.=J.  of  $50=$10,  and 

9.  |  f.=4  times  $10=$40=value  of  first  watch. 

10.  f  s.=f  f.+$30t=f  of  $40+$30=$90=value  of  the  sec- 
ond watch. 
TTT       .    \  $40=value  of  first  watch,  and 
lli<     V    I  $90=value  of  second  watch. 

(  White's  Comp.  Arith.,  p.  248,  prob.  60.) 

I.  At  the  time  of  marriage  a  wife's  age  was  -|  of  the  age  of 
her  husband,  and  10  years  after  marriage  her  age  was 
Y7^  of  the  age  of  her  husband  ;  how  old  was  each  at 
the  time  of  marriage  ? 

1.  |=husband's  age  at  the  time  of  marriage.     Then 

2.  |=wife's  age  at  the  time  of  marriage. 

3.  f+10  years=husband's  age  10  years  after  marriage. 

4.  f +10  years— wife's  age  10  years  after  marriage.     But 

5.  T7u+7  years— y7^  of  (f+10  years )=wife's  age   10  years 
j  i  after  marriage,  by  second  condition  of  the  problem. 

'•  yVK  years=f+10  years.     Whence 
f=10  years — 7  years,  or 

3  years.  [of  marriage. 

=10  times  3  years=30  years=husband's  age  at  time 
or  T6Q,=6  times  3  years=18  years=wife's  age  at  the 
time  of  marriage. 


ANALYSIS.  161 

(  30  years=husband's  age  at  time  of  marriage,  and 
'   '  '      (18  years=wife's  age  at  time  of  marriage. 

(  White's  Comp.  A.,  p.  2^1,  prob.  35.) 

I.  Ten  years  ago  the  age  of  A  was  f  of  the  age  of  B,  and 
ten  years  hence  the  age  of  A  will  be  f  of  the  age  of  B  ; 
find  the  age  of  each. 

1.  |— B's  age  10  years  ago.     Then 

2.  |=A's  age  10  years  ago. 

3.  ^+10  years— B's  age  now,  and 

4.  f+10  years=A's  age  now. 

5.  |+20  years— B's  age  10  years  hence,  and 

6.  f+20  years=A's  age  10  years  hence.  [hence. 

7.  5.  Of   (|_|_20   years)=f+16f  years=A's  age  10   years 

8.  /.  f +16f  years=£+20  years  ;  whence 

9.  5. — 1=20  years — 16|  years,  or 

10.  -^=3^  years,  and 

11.  j-f=12  times  3iJ-  years=40  years=B's  age  10  years  ago. 

12.  |=T9^=9  times  3£  years=30  years=A's  age   10  years 

ago. 

13.  ...  11+10  years=50  years=B's  age  now,  and 

14.  ^2  I  10  years=40  years=A's  age  now. 


TTT  i  &Q  years=B's  age,  and 

Ui'  '*•     (40  years=A's  age. 


I.  Two  men  start  from  two  places  495  miles  apart,  and 
travel  toward  each  other ;  one  travels  20  miles  a  day, 
and  the  other  25  miles  a  day  ;  in  how  many  days  will 
they  meet? 

1.  •J=number  of  days. 

2.  20  mi.=distance  first  travels  in  1  day. 

3.  f  X20  mi.— distance  first  travels  in  f  days. 

4.  25  mi.=distance  second  travels  in  1  day. 

TT  )&•  f  X25  mi.=clistance  second  travels  in  f  days. 

6.  .'.  f  X20  mi.+lX25  mi.=f  X(20  mi.+25  mi.)=distance 

both  travel. 

7.  495  mi.=di stance  both  travel. 

8.  .-.  (20  mi.+25  mi.)x|=(45  mi.)Xf=495  mi.     Whence 
f =495-1-45=1  l=number  of  days. 

III.      .-.  They  will  meet  in  11  days. 

Second  solution. 

rl.  20  miles=distance  first  travels  in  a  day. 

TT  J  2.  25   miles=distance  second  travels  in  a  day. 

3.  .'.  45  miles=distance  both  travel  in  a  day.  [days. 

4.  .-.  495  miles=distance    both    travel    in    495-7-45,  or    11, 


162  FINKEL'S  SOLUTION  BOOK. 

III.     .-.  They  will  meet  in  11  days. 

Third  solution — the  one  usually  given    in  the  schoolroom. 

20+25=45)495(11  days. 
45 
45 
45 

I.     Find  a  number  whose  square  root  is  25  times  its  cube  root. 

1.  -|=square  root  of  the  number.     Then 

2.  !xf=tne  number,  because  the  square  root X the  square 

root  equals  the  number. 

3.  |— the  cube  root  of  the  number.     Then 

4.  fXf  Xf=the  number.     But 


II. 


5.  |=5X(f)-      Hence,  squaring  both  sides, 

6.  fxf=25X(|Xf).     But 

7.  f  xf=the  number,  and 

8.  f  Xf  Xf=the  number. 

9-  .'.  f  Xf  XS=25X(|Xf  )•     Dividing  by  (f  Xf  ), 
10.  f=25. 
[11.  ...  (f  )3=253=15625. 

III.      .-.  The  number  is   15625.  (/?.  H.  A.,  p.  367,prob.  14.) 


I.  A  man  bought  a  horse,  saddle  and  bridle  for  $150  ;  the 
cost  of  the  saddle  was  £  of  the  cost  of  the  horse,  and  the 
cost  of  the  bridle  was  -^  the  cost  of  the  saddle;  what  was 
the  cost  of  each? 

1.  |"f  =cost  of  the  horse.     Then 

2.  •ff=^  of  ^|=cost  of  the  saddle,  and 

3.  Y^=\  of  T2Y=cost  of  the  bridle. 

4.  H=!f+T22+TV=co 
IL<5.  $150=cost  of  all. 


and 

7.  J-^ySr  of  $150=$10=cost  of  bridle. 

8.  H=12  times  $10=$120=cost  of  horse. 
19.  T\=2  times  $10=$20=cost  of  saddle. 

(  $10=cost  of  the  bridle, 
III.  .'.     ]  $20=cost  of  the  saddle,  and 
(  $120=cost  of  the  horse. 

(  White's  Comp.  A.,  p.  241,  prol>.  S9.) 


ANALYSIS. 


163 


I. 


HI. 


A  and  B  perform  T9^  of  a  piece  of  work  in  2  days,  when, 
B  leaving,  A  completes  it  in  -J  day;  in  what  time  can 
each  complete  it  alone? 

1.  T9^=part  A  and  B  do  in  2  days. 

2.  ^=4  °f  A=Part  A  and  B  do  in  1  day. 

3.  -fj — rV=TiF— Part  left  after  B  quits,  and   which   A  com- 

pletes in  ^  day. 

4.  j^=4p=part  A  can  do  in  1  day. 

5.  .'.  -|=part  A  can  do  in  -!—~|=5  days. 

7.  .'•  ^=part  B  can  do  in  |  ;  |,  or  4,  days. 

SA  can  do  the  work  in  5  days,  and 
B  can  do  the  work  in  4  days. 

(  White's  Comp.  A., p.  280,  prob.  193.) 


II. 


I.  A  and  B  can  do  a  p.iece  of  work  in  12  days,  B  and  C  in  9 
days,  and  A  and  C  in  6  days;  how  long  will  it  take 
each  alone  to  do  the  work? 

1.  12  days— time  it  takes  A  and  B  to  do  the  work. 

2.  /.  T1-2-=part  they  do  in  1  day. 

3.  9  days=time  it  takes  B  and  C  to  do  the  work. 

4.  /.  ^— part  they  do  in  1  day. 

5.  6  days=time  it  takes  A  and  C  to  do  the  work. 

6.  /.  ^=part  they  do  in  1  day. 

7    ...  _i__|l|_|_i.==:^3=part  A  and  B,  B  and  C,  and  A  and  C 
do  in  1  day— twice  the  work  A,  B,  and  C  do  in  1  day. 

8.  .*.  Tf=|  of  £f=part  A,  B,  and  C  do  in  1  day. 

9.  i| — ^— y-^-^part  A,  B,  and   C   do  in   1   day — part  B 

and  C  do  in  1  day=part  C  does  in  1  day. 

10.  |f=part  C  does  in  f|-:-TV  or  10f  days. 

11.  T| — ^=T5^^=part  A,  B,  and  C  do  in  1  day — part  B  and 

C  do  in  1  day=part  A  does  in  1  day. 

12.  T|=part  A  does  in  -J|-7-y5^=:14-|  days. 

13.  II — i.— ^i.^part  A,  B,  and  C  do  in  1  day — part  A  and 

C  do  in  1  day=part  B  does  in  1  day. 

14.  Tf==part  B  does  in  ^-^=72  days 

(  14|-  days=time  it  takes  A, 
III.  .-.     <  72  days=time  it  takes  B,  and 
(  10^-  days=time  it  takes  C. 

(  White's  Comp.  A.,  p.  W^prob.  280.) 

I.  The  head  of  a  fish  is  8  inches  long,  the  tail  is  as  long  as 
the  head  and  £  of  the  body-flO  inches,  and  the  body  is 
as  long  as  the  head  and  tail  ;  what  is  the  length  of  the 
fish? 


164  FINKEL'S   SOLUTION   BOOK. 

1.  § =length  of  body. 

2.  8  in.=length  of  head. 

3.  |  1.  of  b.+lO  in.+8  in.=£  1.  of  b.+18  in.=length  of  tail, 

4.  |  1.  of  b.=length  of  head-f-length  of  tail. 


II. 


5.  .-.  f  1.  of  b.=(4  1.  of  b.+18  in.)+8  in.=i  1.  of  b.+26  in 


Whence 

6.  I  1.  of  b.—  i  1.  of  b.=J-  1.  of  b.=26  in. 

7.  .'.  f  1.  of  b.,  or  length  of  body  ,=2  times  26  in.=52  in. 

1.  of  b.+18  in.=26  in.+18  iri.=44  in.=length  of  tail. 
19.   /.  52  in.+44  in.+8  in.s=104  in.=length  of  the  fish. 

III.      .-.  The  length  of  the  fish  is  104  inches. 

I.  Henry  Adams  bought  a  number  of  pigs  for  $48  ;  and 
losing  3  of  them,  he  sold  |-  of  the  remainder,  minus  2, 
for  cost,  receiving  $32  less  than  all  cost  him;  required 
the  number  purchased. 

1.  f=remainder  after  losing  3.     Then 

2.  f-j-3=number  at  first. 

3.  f  of  r. — 2=number  sold. 

4.  $48— $32=$  16= what  was  received  for  f  of  r.— 2. 

5.  $8=£  of  $16=what  was  received   for  -J-  of  (fofr.— 2), 
j  •          or  -|  of  r. — 1. 

6.  $24=3  times   $8=what  was  received    for  3   times  (-£  of 

r— l)=|ofr.— 3. 
.  $48— $24=$24=what  (f  of  r.+3)— (f  of  r.— 3),  or  <3 

pigs  cost. 

4=£  of  $24=what  1  pig  cost. 
.  $48=what  48-T-4,  or  12,  pigs  cost. 

III.     .-.  He  bought  12  pigs. 

(Brooks'  Int.  A.,  f.  164,  Prob-  &•) 

I.  A  bought  some  calves  for  $80;  and  having  lost  10,  he  sold 
4  more  than  §  of  the  remainder  for  cost  and  received 
$32  less  than  all  cost;  required  the  number  purchased. 

1.  |=remainder  after  losing  10.     Then 

2.  f-|-10=number   purchased. 

3.  f  of  r.-|-4=number  sold.  [cost. 

4.  $80— $32=$48=cost  of  f  of  r.+4,   since  they  sold  at 
TT  I  5.  $24=i  of  $48=cost  of  |  of  (f  of  r.+4)=J  of  r.+2. 

^6.  $72=3   times   $24=cost  of  3   times  (i  of  r.+2)=-|  of 
r.-f-6.  [cost. 

7.  .'.  $80— $72=$8=what  (f  of  r.+10)— (f  of  r.+6),  or  4 

8.  $2=^  of  $8=what  1  cost. 

9.  $80=what  80-r-2,  or  40  cost. 

III.     .-.  He  bought  40  calves. 

(Brook's  Int.  A.,  p.  164.,  prob.  10.) 


ANALYSIS.  165 

A  lost  f  of  his  sheep;  now  if  he  finds  5  and  sells  f  of 
what  he  then  has  for  cost  price,  he  will  receive  $18; 
but  if  he  loses  5  and  sells  f  of  the  remainder  for  cost 
price,  he  will  receive  $6;  how  many  sheep  had  he  at 
first? 

1.  J=  the  number  of  sheep  he  had  at  first. 

2.  f=  the  number  he  lost. 

3.  -| — f=f  >  the   number  he  had  after  losing  ^. 

4.  f-f-5=  the  number  he  had  after  finding  5. 

5.  fof  (f+5)=^+3,   the  number  he  sold. 

6.  | — 5=  the  number,  had  he  lost  5. 

7.  | of (| — 5)=^ — 3,  the  number  he  would  have  sold. 

8.  $18=what  (-2VI-3)  sheep  cost. 

9.  $6=  what  (-£% — 3)  sheep  cost. 

10.  ,-.  $12=$18— $6=what  (-fg+S)  sheep— (^—3)  sheep, 

or  6  sheep  cost. 

11.  $2=£  of  $12=  what  1  sheep  cost. 

12.  $18=  what  18-7-2,  or  9  sheep  cost.  But 

13.  $18=  what  (-2\- (-3)  sheep  cost. 

14.  .-.  ^\+3  sheep  =  9  sheep,  or 

15.  ^=6  sheep. 

16.  y^=  J  of  6  sheep=l  sheep,  and 

17.  ff =26  times  1  sheep  =25  sheep. 
.-.  He  had  25  sheep  at  first. 

(Brooks  Int.  A.,  p.  165,  prob.  15.) 

A  man  bought  a  certain  number  of  cows  for  $200;  had  he 
bought  2  more  at  $2  less  each,  they  would  have  cost 
him  $216;  how  many  did  he  buy? 

1.  $200=cost  of  cows. 

2.  .$216— cost  of  oiig'inal  number  of  cows-j-2  more. 

3.  $216— $200=$16=cost  of  2  cows  at  $2  less  per  head. 

4.  .-  $8=4  of  $16=cost  of  1  cow  at  $2.  less  per  head.  Then 

5.  $S+$2=$10=cost  of  each  cow  purchased. 

6.  $200=cost  of  200-r-lO,  or  20  cows. 

.  .  He  bought  20  cows. 

(Brooks  Int.  A.,  p.  162,  prob.  8.) 

A  person  being  asked  the  hour  of  day,  said,  "the  time 
past  noon  is  ^  of  the  time  past  midnight;"  what  was 
the  hour? 

1.  !=time  past  midnight. 

2.  ^=time  past  noon. 

3.  ...  I — K^i—time  from  midnight  to  noon. 

4.  12  hours=time  from  midnight  to  noon. 

5.  .-.  |=12  hours. 

6.  -J=^  of  12  hours=6  hours=time  past  noon. 
III.     It  was  6  o'clock,  P.  M. 


166  FINKEL'S  SOLUTION  BOOK. 

I.     Provided  the  time  past  10  o'clock,  A.  M.,  equals  f  of  the 

time  to  midnight;  what  o'clock  is  it? 
'1.  |=time  to  midnight.     Then 

2.  |=time  past  10  o'clock. 

3.  |-{-f =J=time  from  10  o'clock  to  midnight. 
II.<J  4.   14  hours=time  from  10  o'clock  to  midnight. 

5.  .*.  ^=14  hours. 

6.  £=4  of  14  hours=2  hours,  and  [o'clock  P.  M. 

7.  f=3    times  2  hours=6   hours,  time   past   10  o'clock=4 
III.  '  /.  It  is  4  o'clock,  P.  M. 

I.     At  what  time  between  3  and  4  o'clock  will   the   hour  and 
minute  hands  of  a  watch  be  together? 

1.  -f=distance  the  h.  h.  moves  past  3.     Then 

2.  2^=12xf— distance  the  m.  h.  moves  past  12. 

3.  2/ — -f=2^2=distance  the  m.  h.  gains  on  the  h.  h. 


II. 


4.  15  min.=distance  the  m.  h.  gains  on  the  h.  h 


5.  ...  22=15  min 

6.  -J=^2  °f  15  min.— |f  min.  [past  12. 

7.  2^=24  times  -J-|  min.— 16^-  min.=distance  m.  h.  moves 
III.  *  .'.  It  is  16^i  min.  past  3  o'clock. 


Remark. — In  problems  of  this  kind,  locate  the  minute  hand  at  12  and  the 
hour  hand  at  the  first  of  the  two  numbers  between  which  the  conditions  of 
the  problem  are  to  be  satisfied.  Thus  in  the  above  problem,  at  3  o'clock 
the  minute  hand  is  at  12  and  the  hour  hand  at  3. 

The  minute  hand  moves  over  60  minute  spaces  while  the  hour  hand 
moves  over  5  minute  spaces.  Hence  the  minute  hand  moves  12  times 
as  fast  as  the  hour  hand.  Since  at  3  o'clock  the  minute  hand  is  at  12 
and  the  hour  hand  at  3,  and  the  minute  hand  moves  12  times  as  fast  as  the 
hour  hand,  it  is  evident  that  the  minute  hand  will- overtake  the  hour  hand 
between  3  and  4.  So  we  let  f=distance  the  hour  hand  moves  past  3  until 
it  is  overtaken  by  the  minute  hand.  But  since  the  minute  hand  moves  12 
times  as  fast  as  the  hour  hand,  while  the  hour  move  f,  the  minute  hand 
moves  12  times  f ,  or  ^4.  Now  the  minute  hand  has  moved  from  12  to  3-f|, 
or  15  minutes-)-!,  Hence  the  minute  hand  has  gained  15  minutes  on  the 
hour  hand.  It  has  also  gained  224 — $,  or  %?.  .'.  \2=15  minutes. 

In  solving  any  problem  of  this  nature,  first  locate  the  hands  as  previously 
stated,  and  then  ask. yourself  how  far  the  minute  hand  must  move  to  meet 
tke  conditions  of  the  problem,  if  the  hour  hand  should  remain  stationary. 

I.     At  what  time  between  6   and   7   o'clock   will   the   minute 
hand  be  at  right  angles  with  the  hour  hand? 

1.  |-=distance  h.  h.  moves  past  6. 

2.  V=12  times  f=distance  m.  h.  moves  past  12. 

3.  .*.  2^ — f=2^2=di stance  m.  h.  gains  on  h.  h. 

4.  15  min.  or  45  min.=distance  m.  h.  gains  on  the  h.  h. 

5.  .•.  2^2=15  min.  or  45  min. 

6.  %==•£%  of  15  min.  or  -^  of  45  min.=!~f  min.   or  2-^-  min. 

7.  2^=24    times   ^-J  min.  or  24  times  2-^   min.=16T4T  min. 

or  49yT  min. 

III.     .-.  The  minute  hand  will  be  at  right  angles  with  the  hour 
hand  at  16^  min.  or  49^  min.  past  6  o'clock. 


ANALYSIS. 


167 


Explanation. — Locate  the  minute  hand  at  12  and  the  hour  hand  at  6.  Now 
if  the  hour  hand  had  remained  stationary  at  6,  the  minute  hand  would  have 
to  move  to  3  or  9,  i.  e.,  it  would  have  to  gain  15  min.  or  45  min.  While  the 
minute  hand  is  moving  to  3  the  hour  hand  is  moving  from  0.  So  the  min- 
ute hand  must  move  as  far  past  3  as  the  hour  hand  moves  past  6.  Or  while 
the  minute  hand  is  moving  to  9  the  hour  hand  is  moving  past  6.  So  the 
minute  hand  must  move  as  far  past  9  as  the  hour  hand  is  past  6.  .'.  The 
minute  hand  must  gain  15  minutes  in  the  first  case  and  45  minutes  in  the 
second. 

I.     At  what  time  between  2  and  3   o'clock   are   the   hour  and 
minute  hands  opposite? 

1.  |=distance  hour  hand  moves  past  2.     Then 

2.  2^=distance   the   minute   hand    moves  past    12,   in   the 

same  time.  [hand. 

3.  .'.  2-% — f=2or2=distance  minute  hand  gained  on   the  hour 
!!.<!  4.  40  min.=distance  the  minute  hand  gained  on   the  hour 

hand. 

5.  .*.  2Y2=40  min. 

6.  •J=^£  of  40  mim.=l-j\-  min.,  and 

7.  2^=24  times  1T9T  min.=43i7y  min. 

III.     .•.  It    is    43T7T  min.    past    2    o'clock  when   the   hands   are 
opposite. 

Explanation. — Locate  the  minute  hand  at  12  and  the  hour  hand  at  2.  Now 
if  the  hour  hand  remained  stationary  at  2,  the  minute  hand  would  have  to 
move  to  8  or  over  40  minutes  in  order  to  be  opposite  the  hour  hand.  But 
while  the  minute  hand  is  moving  to  8,  the  hour  hand  is  moving  from  2.  So 
the  minute  hand  must  move  as  far  past  8  as  the  hour  hand  is  past  2.  Since 
\  is  the  distance  the  hour  hand  moves  past  2,  f  must  be  the  distance  the 
minute  hand  must  move  past  8.  Hence  the  distance  the  minute  hand 
moves  is  f+40  min.  But  \4=distance  the  minute  hand  moves.  .'.  V=l+ 
40  min.  or  ^=40  min.  as  shown  in  step  5. 


I. 


II. 


At  what  time  between   3   and   4   o'clock   will   the  minute 
hand  be  5  minutes  ahead  of  the  hour  hand? 

1.  |—  distance  hour  hand  moves  while  the  m.  h.  is  moving 

to  be  5  min.  ahead.  [moves  -|  . 

2.  2^=12  Xf=distance  minute  hand  moves  while  the  h.  h. 

3.  /.  2^  —  f=2T2=distance  gained  by  the  minute  hand. 

4.  15  min.-}-5  mim.=:20  min.=distance  gained  by  the  m.  h. 

5.  .-.  %2=20  min. 

6.  i=^  of  20  min.=4-f  min. 

7.  V==24  times  TT  min—21-jSr  min. 


III.     /.It  is  21T9T  min.  past  3  o'clock. 

Explanation.  —  Locate  the  minute  hand  at  12  and  the  hour  hand  at  3. 
Now  if  the  hour  hand  remained  stationary  at  3,  the  minute  hand  would 
have  to  move  to  4  in  order  to  be  5  min.  ahead.  But  while  the  minute  hand 
is  moving  to  4  the  hour  hand  is  moving  from  3.  Hence  the  minute  hand 
must  move  as  far  past  4  as  the  hour  hand  moves  past  3.  But  the  hour  hand 


168 


FINKEL'S   SOLUTION  BOOK. 


moves  |  past  3;  hence,  the  minute  hand  must  move  f  +5  min.  past  4,  in  all, 
|+20  min.  Hence,  the  minute  hand  gains  (|+20  min.)  —  |=20  min.  on  the 
hour  hand. 

Remark.  —  We  always  find  \*,  the  distance  the  minute  hand  moves,  for  it 
indicates  the  time  between  any  two  consecutive  hours.  The  hour  hand 
indicates  the  hour. 


I. 


II. 


At  what  time  between  4  and  5  o'clock  do  the   hands  of  a 
clock  make  with  each  other  an  angle  of  45°  ? 

1.  |=distance  the  hour  hand  moves  past  4. 

2.  2^=distance  the  minute  hand  moves  past  12. 

3.  .'.  224  —  f=2Y2=  distance   the   minute   hand   gains   on  the 

hour  hand. 


III. 


4.  12£  min.  or  27-J  min.=distance  gained   by  minute  hand. 

5.  /.  2Y2=12£  min.  or  27£  min.  [min. 

6.  -|=-j-2    of  12-J  min.  or  •£%  of  27-J  min.=J|-   min.   or   1J 

7.  2^=24    times    -|^  min.  or  24  times   1^  min.=1313r  min. 

or  30  min. 

.*.  At  13T7T  min.  past  4  or  30  min.  past  4,  the  hands   make 
an  angle  of  45°  with  each  other. 

Explanation. — Locate  the  minute  hand  at  12  and  the  hour  hand  at  4.  45° 
=i  of  360°.  &  of  60  min.=7$  min.  Hence,  that  the  hands  make  an  angle  of 
45  ,  the  minute  hand  must  be  either  7£  minutes  behind  the  hour  hand  or  7| 
min.  ahead,  Now  if  the  hour  hand  remained  stationary  at  4,  the  minute 
hand  would  have  to  move  over  12£  min.  or  2|  min.  past  2.  But 
while  the  minute  hand  is  moving  this  distance,  the  hour  hand  is  moving 
past  4.  Hence,  the  minute  hand  must  move  as  far  past  2£  min.  past  2  as  the 
hour  hand  moves  past  4,  z'.  e.,  the  minute  hand  moves  f+124  min.  Hence, 
it  gains  (f-f-12£  min.) — 1=12|  min.  The  reasoning  for  the  second  result  is 
the  same  as  for  the  first. 

I.     At  what  time  between  4  and  5  o'clock  is  the  minute  hand 
as  far  from  8  as  the  hour  hand  is  from  3  ? 

1.  -|=distance  the  hour  hand  moves  past  4. 

2.  2^=12  times  f=distance  minute   hand   moves  past 

12  in  the  same  time. 

3.  .'.  V+f=  V=distance  both  move. 

4.  35  min.— distance  both  move. 

5.  /.  2^6=35  min. 

6.  •^=-5-$  of  35  min.=l^  min. 

7.  V=?4  times  1-fc  min.=32T\  min. 

1.  f=distance  the  h.  h.  moves  past  4. 

2.  2Y4=distance  minute  hand  moves  past  12. 

3.  .'.  2-£ — 2f:=?22=di stance  the  minute  hand  gains. 

4.  45  min.=distance  the  minute  hand  gains. 

5.  .•.  2^2=45  min. 

6.  -J=^-  of  45  min.=2^y  min. 

7.  2Y4=24  times  2^  min.=4911r  min. 

It  is  32T\  min.  or  49yT  min.  past  4  o'clock. 

(7?.  H.  A.,  p.  403,prob.  40.) 


II. 


A. 


B. 


III. 


ANALYSIS. 


16& 


Explanation.  —  This  problem  requires  two  different  solutions.  Locate  the 
minute  hand  at  12  and  the  hour  hand  at  4.  The  hour  hand  is  now  6  min- 
utes from  3.  If  the  hour  hand  remained  stationary,  the  minute  hand  would 
have  to  move  to  7  to  be  5  minutes  from  8.  But  while  the  minute  hand  is 
moving  to  7,  the  hour  hand  is  moving  past  4.  Hence  the  minute  hand  must 
stop  as  far  from  7  as  the  hour  hand  moves  past  4;  «.  <?.,  if  the  hour  hand 
moves  f  past  4  the  minute  hand  must  stop  f  from  7.  Then  the  hour  hand 
will  be  5  minutes+|  from  3  and  the  minute  hand  will  be  f-j-5  minutes  from 
8.  While  the  hour  hand  moved  f  ,  the  minute  hand  moved  35  min.  —  $  /.  ^ 
=35  min.  —  $,  whence  ^=35  min.  .*.  35  min.=distance  they  both  move. 
The  second  part  has  been  explained  in  previous  problems. 


I. 


At  what  time  between  5  and  6  o'clock  is  the  minute  hand 
midway  between  12  and  the  hour  hand?  When  is  the 
hour  hand  midway  between  4  and  the  minute  hand? 


rA. 


II. 


B. 


!=distance  the  hour  hand  moves  past  5. 
2^=distance  the  minute   hand   moves  in   the   same 

time. 

f  +25  min.=distance  from  12  to  the  hour  hand. 
-J-  of  (f-f-25  min.)=-J-[-12-J-  min.=distance   minute 

hand  moves. 


III. 


5. 
6. 

7. 

8 
9. 

A. 
B. 


mn- 

i=-^g-  of  12-j-  min.=Jf  min. 
V=24  times  ff  min.=13^  min. 
|-=distance  the  hour  hand  moves  past  5. 
2^=distance   the  minute   hand    moves  in  the   same 

time 

f+5  min.=distance  the  hour  hand  is  from  4. 
Y+10  min.  =2  times  (f+5  min.)=distance  the  min- 

ute hand  is  from  4,  since  the  hour  hand  is  midway 

between  it  and  4. 
20  min.+(f+10  min.)=f-f-30  min.=distance  the 

the  minute  hand  is  from  12. 


==__      mm.,  or 


of  30  min.=l£  min. 
=24  times  H  min.=36  min. 
It  is  13-^g-  min.  past  5  o'clock. 
It  is  36  min.  past  5  o'clock 


. 
(R.  H.  A.,  p.  403,prob.  41.) 


Explanation.  —  Locate  the  minute  hand  at  12  and  the  hour  hand  at  5.  If 
the  hour  hand  remained  stationary,  the  minute  hand  would  have  to  move 
over  \  of  25  minutes,  or  12£  minutes.  But  while  it  is  moving  over  12$ 
minutes,  the  hour  hand  is  moving  past  4.  Hence,  the  minute  hand  will 
have  to  move  12£  minutes+^  of  the  distance  the  hour  hand  moves  past  4. 
Hence  V=i~l~12|  minutes,  as  shown  by  step  5  of  A.  In  B,  if  the  hour  hand 
remained  stationary,  the  minute  hand  would  have  to  move  over  30  minutes, 
i.  e.,  to  6,  that  the  hour  hand  may  be  midway  between  it  and  4.  But  while 
the  minute  hand  is  moving  to  6  the  hour  hand  is  moving  past  4.  Hence 
the  minute  hand  must  move  twice  as  far  past  6  as  the  hour  hand  moves  past 


170 


FJNKEL'S   SOLUTION  BOOK. 


4.  But  |=distance  the  hour  hand  moves  past  4;  hence,  £=distance  the  min- 
ute hand  moves  past  6.  Hence,  1+30  minutes=distance  the  minute  hand 
moves.  .'.  ^*—  f+30  minutes,  as  shown  by  step  6  of  B. 


I. 


II.< 


At  what  time  between  3  and  4  o'clock  will  the  minute 
hand  be  as  far  from  12  on  the  left  side  of  the  dial  plate 
as  the  hour  hand  is  from  12  on  the  right  side? 

distance  the  hour  hand  moves  past  3. 
=12  times  |=distance  the  minute  hand  moves  in  the 
same  time. 

3.  2^4_j_|==2_6_:^istance  they  both  move. 

4.  45  min.=distance  they  both  move. 

^=45  min. 

-^  of  45  min.=l|-|  min. 

=24  times  144  min.=41-A-  min. 


5. 
6. 

7. 

III.     .-.  It  is  41^  min.  past  3. 

Explanation. — Locate  the  minute  hand  at  12  and  the  hour  hand  at  3.  If 
the  hour  hand  remained  stationary,  the  minute  hand  would  have  to  move  to 
9  to  be  as  far  from  12  on  the  left  side  of  the  dial  plate  as  the  hour  hand  is 
from  12  on  the  right.  But  while  the  minute  hand  is  moving  to  9,  the  hour 
hand  is  moving  past  3.  Hence,  the  minute  hand  must  stop  as  far  from  9  as 
the  hour  hand  moves  past  3.  Hence,  it  is  evident,  they  both  move  45 
minutes. 


i. 


II. 


A  man  looked  at  his  watch  and  found  the  time  to  be  be- 
tween 5  and  6  o'clock  Within  an  hour  he  looked 
again,  and  found  the  hands  had  changed  places.  What 
was  the  exact  time  when  he  first  looked? 

|-=distance  m.  h.  was  ahead  of  h.    h.,   or   the  dis- 
h.    moved,    since   it  changed    place 


(8.) 


III.     .'. 


tance  the  h. 

with  the  m.  h.  [the  two  observations. 

2^=distance  the  m.  h.  moved  in  the   time  between 
•"•  V44-f:=V=distance  they  both  moved. 
60  mm.=distance  they  both  moved. 
.-.  2^=60  min. 

-i=Jg.  of  60  min.=2-j^  min.  [ahead  of  h.  h. 

f=2  times  2T\  min.=4T%  min.=distance  m.  h.  was 
1.  |^=distance  h.  h.  was  past  5,  at  time  of  first  obser- 
vation.    Then  [servation. 
2^=distance  m.  h.  was  past  12  at  time  of  first  ob- 
25    min.-}-f+4T%   min.=f+291^=  distance   m.   h. 

was  past  12  at  time  of  first  observation. 
4. 
5. 

6.  -j— ^2  of  29T83-  rrin.=l^-  min. 
-7.  224— 24  times  1-^-  min.=32T%  min. 

It  was  32T43  min.  past  5  o'clock. 


ANALYSIS. 


171 


Explanation.  —  It  is  clear  that  the  minute  hand  was  ahead  of  the  hour 
hand  at  the  time  of  the  first  observation,  or  else  they  could  not  have  ex- 
changed places  \vithin  an  hour.  Now,  we  call  the  distance  from  the  point 
where  the  hour  hand  was  located  at  first  lothe  point  where  the  minrte  hand 
was  located  first,  -|.  But  in  the  mean  time  the  hour  hand  has  moved  lo  the 
position  occupied  by  the  minute  hand  and  the  minute  hand  ha«  moved  on 
around  the  dial  to  "the  position  occupied  by  the  hour  hand,  /.  c.,  the  hour 
hand  has  moved  |  and  the  minute  12  times  f,  or  V-  Hence,  they  both 
moved  ^6.  They  both  moved  60  minutes  since  the  hand  moved  on  around 
the  dial  to  the  position  occupied  by  t'he  hour  hand  and  the  hour  hand  mov- 
«d  to  the  position  occupied  by  the  minute  hand.  .•  \6—  60  min.  as  shown 
in  step  (5.)  The  remaining  part  of  the  solution  has  been  explained  in  pre- 
vious problems. 

At  a  certain  time  between  8  and  9  o'clock  a  boy  stepped 
into  the  schoolroom,  and  noticed  the  minute  hand  be- 
tween 9  and  10.  He  left,  and  on  returning  within  an 
hour,  he  found  the  hour  hand  and  minute  hand  had  ex- 
changed places.  What  time  was  it  when  he  first  en- 
tered, and  how  long  was  he  gone? 

((1.)   f=distance  m.  h.  was  ahead  of  the  h.    h.   or  dis- 
tance it  moved.  [J. 

2^4=distance  m.  h.  moved  while  the  h.    h.  moved 
2_4_|_|=  2^6  ^13.^^  both  moved. 
(4  )  60  min  =distance  both  moved. 
^=60  min. 

YV  °f  60  min.=2T\  min.  [was  ahead 

(7.)  f=2   times    2T\   min.=4-£%  min.=distance   m.  h. 
rl.   |=distance  h.  h.  moved  past  8- 

2.  21£=distance  m.  h.  moved  in  same  time. 

3.  40    min+|+4T%    min.=|+441^    min.  =  dis- 

tance m.  h.  moved  to  be  4T%  min.  ahead. 


rA. 


II. 


(2.) 
(3.) 


(5.) 
(6.) 


(8.)^  4. 
5. 

6.  -f— 2T  of  44T8-g-  min.=2-i£-g-  min.  [past  8. 

7.  2T4=24    times    2T|7   min.  =  48^  min.=time 

1.  2/>— di stance  thty  both  moved. 

2.  60  min. —distance  they  both  moved. 
B.                3.  .-.  2T6=6C   min. 

4.  ^=2if  of  60  min.=2T\  min.  [was  gone. 

5.  \4— 24  times   2T%   min.=55T\   min.=time   he 

(  A.  It  was  48T9¥6¥  min   past  8  o'clock  when  he  first  en- 
<  tered  school  room. 

(  B.  He  was  gone  55T\  min. 

Suppose  the  hour,  minute,  and  second  hands  of  a  clock 
turn  upon  the  same  center,  and  are  together  at  12 
o'clock;  how  long  before  the  second  hand,  hour  hand, 
and  minute  hand  respectively,  will  be  midway  between 
the  other  two  hands  ? 


172 


FINKEL'S   SOLUTION   BOOK. 


'A. 


II. 


B. 


.C. 


1.  |=distance  the  hour  hand   moves   past  12.     Then 

2.  ^—distance  the  minute  hand  moves  past   12,  and 

3.  i_4^4_o=720  times  |=distance 
the  second  hand  moves  past 
12. 

4.  i_4_4_o_2_4  ^  i_4_i_6  =  distance 

from  the  minute  hand  to  the 
second  hand. 

5.  i^u  —  y ==uyj  ==  distance 
from  the  second  hand  to  the 
hour  hand. 

2^ — |-  =  2^2  =  distance    from 
the  hour  hand  to  the  second  hand. 
1-4Y1-6  +  1^1~H~¥=2~TJ^=distance   around   the  dial. 
60  seconds=distance  around  the  dial   as   indicated 
by  one  revolution  of  the  s.  h. 


6. 

7. 
8. 

9. 
10. 
11.  MP==I440  times 


=  °f  60  S6C— 


sec.=30T\Vr   sec.  =  time 


when  s.  h0  is  midway  between  the  h.  h.  and  m.  h. 
^=distance  the  hour  hand  moves   past   12.     Then 
^distance  the  minute  hand  moves  past   12,  and 

distance    the    second 
hand  moves  past  12. 

4.  *-£ — f=2Y2=distance  from  h. 
h.  to  in.  h. 

5.  2Y2=distance  from  s.  h.   to  h. 
h.,  because  the  h.  h.   is  mid- 
way between  them.          [12. 

6.  2^ — f=distance  from  s.  h.  to 

7.  1-J2-4-°+y)  =  1-4#-°  =  distance 
around  the  dial. 

8.  60  sec.=distance  around  the  dial. 

9.  /.  1 46-0=60  sec. 


TIQ.A. 


of  ^0  sec.=^  sec. 
11.   Uyy^^o  times  T3¥  sec.=59|f  sec.=time  when 

the  h.  h.  is  midway  between  the  s.  h.  and  m.  h. 
1    |-=distance  h.  h.  moves  past  12.     Then 

2.  2^=distance  m.  h.  moves  past  12,  and 

3.  1-^-°=distance   s    h.   moves 
past  12.  [h.  to  s.  h. 

4.  224  —  f=2Y2=distance  from  h. 
5:   ^—distance   from   m    h.  to 

s.  h.  [from  12  to  s   h. 

6.  |-  +  222  +  V  =  V  =  distance 

7.  -JytP—  4_e  =  i_3_9_4  ^  distance 
around  the  dial.  [dial. 

8.  60  sec  =distance  around  the 


ANALYSIS. 


173 


III.  ,\ 


=         Sec. 


=        sec- 


9. 

10. 

11.   14240=  1440  times  ^  sec.=61f|f  sec.=time  past 

12  when  the  m.  h.   will   be   midway   between  the 

h,  h.  and  s.  h. 

»A.  The  second  hand  is  midway  between  h.  h.  and  m. 
h.  at  30T\9Y°7  sec.  past  12.        [at  59f|  sec.  past  12. 

B.  The  hour  hand  is  midway  between  s.  h.  and  m.  h. 

C.  The  minute  hand  is  midway  between  h.  h.    and  s. 


h.  at 


sec.  past  12. 


•'•  From  m  to  .$•  =  Ts—Tm  =  *  Y°—  V  =  1  V6-     And»  b7  the  condi- 
tion of  the  problem,  the  distance  from  m  to  s  =  the  distance  from  m  to  h.   :. 


Explanation.  —  A.  We  represent  the  distance  moved  by  the  hour  hand 
by  |,  =  the  space  Th.  And  since  the  minute  hand  moves  12  times  as  fast 
as  the  hour  hand,  it  moves  ^*.  The  second  hand  moves  60  times  as  fast  as 
the  minute  hand  or  720  times  as  fast  as  the  hour  hand.  From  T  to  h  is  f 
and  from  I  to  m  is  ^*.  .'.  From  h  to  m  is  Tm  —  7V/  =  *f  —  f  =  ^.  From  T  to 
s  is  *  Y°- 
of  t 

from  w  to  ^  =  14§l6  +  1V6-2¥2-     We  nave  seen,  already,  that  the  distance 
from  h  to  m  is  \2.     .'.  The  whole  distance  around  the  dial  is  2{y*2  +  222  =  2*g4. 

B.  From  T  to  k  is  f.     From  T  to  m  is  V-     •'•  From  //  to  m=Tm—Th= 
^*  —  i=V'     By  the  condition  of  the  problem,  the  distance  from  h  to  m=the 
distance  from  .9  to  //.     /.  sT—  T/i==^  —  \=*g.     From   T  around  the  dial  to 
the  right  of  s  is  1^°.     .'.  The  whole  distance  around  the  dial=1V°+^Q= 

1  -CJ5  0 

C.  From  T  to  h  is  f  .     From  T  to  m  is  ^*.     .'.  From  //  to  m=Z£—\=*$-. 
By  the  condition  of  the  problem,  the  distance  from  m  to  .?—  the  distance 
from  h  to  m=  ^.     .'.  From   T  to  s  is  1+^+^=^.     From  T  around  the 
dial  through  T  to  s  is  ^s>.     .'.  The  whole  distance  around  the  dial 


I. 


II. 


III. 


A  sold  to  B  9  horses   and   7 
same   price,  6   horses   and 


cows   for  $300;  to  C,  at  the 
13   cows,  for   the  same  sum; 


what  was  the  price  of  each? 

1.  Cost  of  9  horses-f-cost  of  7  cows=$300.     Then  the 

2.  Cost  of  36  horses-j-cost  of  28   cows=$1200,  by  taking 

4  times  the  number  of  each. 

3.  Cost  of  6  horses-|-cost  of  13  cows=$300.     Then  the 

4.  Cost  of  36  horses+cost  of  78  cows=$1800,  by  taking 

6  times  the  number  of  each.     But 

5.  Cost  of  36  horses+ccst  of  28  cows=$1200. 

6.  .'.  Cost  of  50  cows=$600,  by  subtracting;  and 

7.  Cost  of  1  cow=-^j-  of  $600=$12.     The 

8.  Cost  of  7  cows=7  times  $12=$84. 

9.  .-.  Cost  of  9  horses-=$300— cost  of  7   cows=$300— $84 

=$216.     The 
10.  Cost  of  1  horse=!  of  $216=$24. 

The  cows  cost  $12  apiece,  and 
The  horses  $24  apiece. 


II. 


174  FINKEL'S  SOLUTION  BOOK. 

I.  A  man  at  his  marriage  agreed  that  if  at  his  death  he 
should  leave  onlv  a  daughter,  his  wife  should  have  ^  of 
his  estate,  and  if  he  should  leave  only  a  son  she  should 
have  \.  He  left  a  son  and  a  daughter.  What  fractional 
part  of  the  estate  should  each  receive,  and  what  was 
each  one's  portion,  if  his  estate  was  worth  $6591? 
-1.  ^=daughter's  share. 

2.  ^=wife's  share. 

3.  |=3  times  f=son's  share. 

4.  J_j_|._|_j_i?3=:i_w]loie  estate. 

5.  $6591=whole  estate. 

6.  /.  ^=$6591.  [estate. 

7.  i=T^  of  $6591=$507=daughter's   share,=^g-  of  whole 
|=3  times  $507=$1521=wife's  share,=T\  of  whole  es- 
tate, [tate. 

|=9  times  $507=$4563=son's  share ,=3%  of  whole  es- 

(  $507=Y*g-  of  whole  estate=daughter's  share. 
III.     .-.   ]  $1521=T3¥  of  whole  estate=wife's  share. 
(  $4563=T9^  of  whole  estate=son's  share. 

(Milne's  Prac.  A.,  p.  362,  prob.  74.) 

Note. — For  a  valuable  critique,  by  Marcus  Baker,  U.  S.  Coast  Survey,  on 
this  class  of  problems,  see  School  Visitor,  Vol.  IX.,  p.  186. 

I.  There  is  coal  now  on  the  dock,  and  coal  is  running  on 
also  from  a  shoot  at  a  uniform  rate.  Six  men  can  clear 
the  dock  in  1  hour,  but  11  men  can  clear  it  in  20  min- 
utes ;  how  long  would  it  take  4  men  ? 

1.  |=what  one  man  removes  in  1  hour.     Then 

2.  1g8=6  times  |=what  6  men  remove  in  1  hour. 

3.  f=-j  of  |=what  1  man  removes  in  20  min.,  or  -J-  hour. 

4.  ¥=H  times  |=what  11  men  remove  in  %  hour. 

5.  /.  lf — ¥=y=wnat  runs   on  in   1   hr-~ i  hr.=|  hr. 

Then 
II. J    6.  £=1-^-7-f=what  runs  on  in  1  hour.  [commenced. 

7.  .*.  lf — |=|=what  was  on   the   dock   when   the  work 

8.  |=what  4  men  remove  in  1  hour. 

9.  .'.  | — J=-J=part  of  coal  removed  every  hour,  that  was 

on  the  dock  at  first. 
10.  j=coal  to  be  removed  in  |-s-£=5  hours. 

(R.  H.  A., p.  406, prob.  90.) 
III.     .-.  It  will  take  4  men,  5  hours  to  clear  the  dock. 

Explanation. — ^2— what  6  men  remove  in  1  hr.  and  2e2=what  11  men  re- 
moved in  £  hr.  In  either  case  the  dock  was  cleared.  /.  ^ — ^r^g4— 
amount  of  coal  that  ran  on  the  dock  from  the  shoot  in  1  hr. — £  hr  ,  or  §  hr. 
Hence  in  1  hr.  there  will  run  on,  **---%— 1*.  Since  \  run  on  in  1  hr.  and  ^ 
—the  whole  amount  of  coal  removed  in  1  hr.,  ^2 — |,  or  f  must  be  the 
amount  of  coal  on  the  dock  when  the  work  began.  Since  |=the  amount  4 
men  rerrove  in  1  hr.  and  |— the  amount  that  runs  on  the  dock  in  1  hr.,  | — 
|,  or  |  is  the  part  of  the  original  quantitv  removea  each  hour.  Hence,  if  \ 
is  removed  in  1  hour  |  would  be  removed  in  | -±-\,  or  hours. 


ANALYSIS.  175 

If  12  oxen  eat  up  3£  acres  of  pasture  in  4  weeks,  and  21 
oxen  eat  up  10  acres  of  like  pasture  in  9  weeks  ;  to  find 
how  many  oxen  will  eat  up  24  acres  in  18  weeks. 

1.  10  parts  (say)=what  one  ox  eats  in  a  week.     Then 

2.  120  parts=12XlO  parts=what  12  oxen  eat  in  1  week, 

3.  480  parts=^4x!20  parts=what  12  oxen  eat  in  4  weeks. 

4.  .'.  480  parts=original  grass-|-growth  of  grass  on  3£  A. 

in  4  weeks. 

5.  144  parts=—  of  480  parts=original  grass-|-growth  of 

grass  on  1  A.  in  4  weeks. 

6.  210  parts=21XlO  parts=what  21  oxen  eat  in  1  week, 

7.  1890  parts=9  X210  parts=what  21  oxen  eat  in  9  weeks. 

8.  .'.  1890  parts=original  grass-j-growth   of  grass  on   10 

A.  in  9  weeks. 

9.  189  parts=TL    of   1890  parts=original   grass-fgrowth 

on  1  A    in  9  weeks 

10.  .'.  189  parts — 144    parts=45   parts=growth  on  1  A.  in 

9  weeks — 4  weeks,  or  5  weeks. 

11.  9  parts— i  of  45  parts=growth  on  1  A.  in  1  week. 

12.  36  parts=4x9  parts=growth  on  1  A.  in  4  weeks. 

13.  .-.  144  parts — 36  parts=108  parts=original  quantity  of 

grass  on  1  A. 

14    2592   parts=24xl08  parts=original  quantity  on  24  A. 
15.  216  parts=24x9  parts=growth  on  24  A.  in  1  week. 
16    3888    parts=18x216    parts=growth  on   24   A.  in   18 

weeks. 

17.  /.  2592    parts+3888    parts=6480    parts=quantity    of 

grass  to  be  eaten  by  the  required  oxen. 

18.  180  parts=18X10  parts=what  1  ox  eats  in   18  weeks. 

19.  .'   6480   parts=what  6480-—180,  or  36  oxen  eat  in  18 

weeks. 

III.     .-.  It  will  require  36  oxen  to   eat  the  grass  on  24  A.  in  18 
weeks. 


Note. — This  celebrated  problem  was,  very  probably,  proposed  by  Sir 
Isaac  Newton  and  published  in  his  Arithmetica  Universalis  in  1704.  Dr. 
Artemas  Martin  says,  "I  have  not  been  able  to  trace  it  to  any  earlier  work." 
For  a  full  treatment  of  this  problem  see  Mathematical  Magazine,  Vol.  1, 
No.  2. 


I.  A  man  and  a  boy  can  mow  a  certain  field  in  8  hours,  if 
the  boy  rests  3|  hours,  it  takes  them  9£  hours.  In  what 
time  can  each  do  it? 


176  FINKEL'S   SOLUTION  BOOK. 

1.  9J  hr. — 3f  hr.=5f  hr.=time  they  both  work  together  in 

the  second  case. 

2.  8  hr.=time  it  takes  them  to  do  the  work. 

3.  .-.  -J=part  they  do  in  1  hour. 

4.  -^=ff=5f  times  -J— part  they  do  in  5f  hours. 

8 

IL<[5.  .'.  || — f|=79a=part  the  man  did  in  3|  hours,  while  the 
boy  rested. 

6.  /.  ¥3ff=—  of -^j=part  the  man  did  in  1  hour. 

df 

7.  .-.  |_jj=part  the  man  can  do  in  ^-§-7-^  or  13-^-  hours. 

8.  -£ — ^=£1-5=part  the  boy  does  in  one  hour. 

9.  .•.  |-jj=part  the  boy  can  do  in  f  ^-r-^V  or  20  hours. 
TTT  (  It  will  take  the  man  13-J  hours,  and 

*   /  The  boy  20  hours.  (R.  H.  A., p.  402,prob.  30.) 

I.  Six  men  can  do  a  work  in  4£  days;  after  working  2  days, 
how  many  must  join  them  so  as  to  complete  it  in  3| 
days? 

1.  4£  days=time  it  takes  6  men. 

2.  26  days=6  times  4^-  days=time  it  takes  1  man. 

3.  /.  ^g-— part  1  man  does  in  1  day. 

4.  Yg=$  times  gL=part  6  men  do  in  1  day. 

5.  yV=2  times  T3^=part  6  men  do  in  2  days.  [days. 

6.  Tf — T6^=y7^=part  to  be  done  in  3f  days — 2  days,  or  If 


7.  Kj=T3TT:=IPart  1  man  does  in  1^  days. 


II. 


or  10  men  can  do  in  1-|  days. 
9.  .*.  10  men — 6  men=4   men,  the   number   that    must  join 

them. 

III.     /.  They  must  be  joined  by  4  more  men  that  they  may  com- 
plete the  work  in  3|  days.       R.  H.  A.,  p.  402, prob.  34. 

I.     From  a  ten-gallon   keg   of  wine,  one   gallon   is  drawn  off 
and    the   keg  filled   with  water ;  if  this   is   repeated  4 
times,  what  will  be  the  quantity  of  wine  in  the  keg? 
'1.  J^z^part  drawn  out  each  time. 

2.  T9^=part  that  was  pure  wine  after  the  first  draught. 

3.  y1^  of  yVr=Tir7=:Part:  wine  drawn  off  the  second  draught. 

4.  ^ — r|7y=:T8^=  part  pure   wine    left,   after    the   second 

draught.  [draught. 

5.  yg-  of  y8^  ==  T-| i-Q-  =  part    wine    drawn    off  at   the   third 
y8^ — yl^^y7^9^— part   pure   wine   left   after  the  third 

draught.  [draught. 

TTT  °f  T7Tnnr==rJMir===Par^:  w^ne  drawn  off  at  the  fourth 
TTsV- TW£ff=TWinr=  Part  Pure  wine  left  after  fourth 

draught.  [fourth  draught. 

9.  /.  -j^nnr  °^  10  gal.=6.561  gal.=pure  wine  left  after  the 


II. 


draught. 

9292 

5.  yV  of  (— )  ==—— ==part   wine   drawn    off  at  the  third 

draught. 

6.  (-^)  — ^r=(^}  =part   wine    left    after    the    third 


PROBLEMS.  177 

III.     .*.  There  will  be  6.561  gal.  of  pure  wine   in   the  keg  after 
the  fourth  draught. 

I.     In  the  above  problem,  how  many  draughts   are  necessary 
to  draw  off  half  the  wine? 

1.  T177==part  wine  drawn  off  at  the  first  draught. 

2.  \    —I\=^=part  wine  left  after  the  first  draught. 

Q 

3.  TV  of  T9¥— T-^— — — =part  wine   drawn   off  at  the  sec- 

ond draught. 

92  9    2 

4>  T9iF— ^TV(T=(T)  =part  wine  left  after  the  second 


II. 


draught.     By  induction, 

7.  (T^)n=rpart  wine  left  after  the  nth  draught. 

8.  /.  10(T9Tr)n=number  of  gal.  left  after  the  ^th  draught. 

9.  5=number  of  gal.  left  after  the  n\h  draught. 
10.  .'.  10(T9Tr)n=5,  whence 

H-   (T97y)n=i-     Applying  logarithms, 

12.  n  log.  TV=log.  f 

13.  /.  n  =  log.  i-r-log.  TV=-30103-:-.  T.954243=.301030H- 

.045757=6+. 

III.     /.  In  7  draughts,  half  and  a  little   more   than   half  of  the 
wine  will  be  drawn  off. 

PROBLEMS. 

1.  A  man  bought  a  horse  and  a   cow   for  $100,  and   the  cow 
cost  f  as  much  as  the  horse;  what  was  the  cost  of  each? 

Ans.  horse,  $60;  cow,  $40. 

2.  Stephen   has   10  cents   more   than   Marthia,   and   they  to- 
gether have  40  cents;  how  many  have  each? 

Ans.  Stephen,  25/;  Marthia,  15/. 

3.  A's  fortune  added  to  -J  of  B's  fortune,  equals  $2000;  what 
is  the  fortune  of  each,  provided  A's  fortune  is  to  B's  as  3  to  4? 

Ans.  A's,  $1200;  B's,  $1600. 

4.  If  10  oxen  eat  4  acres  of  grass  in   6   days,   in   how  many 
days  will  30  oxen  eat  8  acres?  Ans.  4  days. 


178  FINKEL'S   SOLUTION  BOOK. 

5.  If  a  5-cent  loaf  weighs  7  oz.  when  flour  is  worth  $6  a  bar- 
rel, how  much  ought  it  weigh  when  flour  is  worth  $7  per  barrel  ? 

Ans. 

6.  A  lady  gave  80  cents  to  some  poor   children;    to   each  boy 
she  gave  2  cents,  and  to  each  girl  4  cents;  how  many  were  there 
of  each,  provided  there  were  three  times  as  many   boys  as  girls? 

Ans.  8  girls;  24  boys. 

7.  Two  men    or  three  boys  can    plow  an  acre  in  ^  of  a  day  ; 
how  long  will  it  take  3  men  and  2  boys  to  plow  it? 

Ans.  J-j  da. 

8.  A  agreed  to  labor  a  certain  time  for  $60,  on   the  condition 
that  for  each  day  he  was  idle  he  should  forfeit  $2,  at   the  expira- 
tion of  the  time  he  received  $30;  how   many    days   did   belabor, 
supposing  he  received  $2  per  day  for  his  labor?      Ans.  22-J  days. 

9.  The  head  of  a  fish  is  4  inches  long,  the   tail   is  as   long  as 
the  head,  plus  -J  of  the  body,  and  the  body  is  as  long  as  the  head 
and  tail ;  what  is  the  length  of  the  fish?  Ans.  32  inches. 

10.  In  a  school  of  80  pupils  there  are  30  girls;  how  many  boys 
must  leave  that  there  may  be  3  boys  to  5  girls?  Ans.  32. 

11.  A  steamboat,  whose  rate  of  sailing   in   still   water  is  12 
miles  an  hour,  descends  a  river  whose  current  is  4  miles  an  hour 
and  is  gone  6  hours;  how  far  did  it  go?  Ans.  32  miles. 

12.  A  man  keeps  72  cows  on  his  farm,  and  for  every   4  cows 
he  plows  1  acre,  and  keeps  1  acre  of  pasture   for  every   6  cows  ; 
how  many  acres  in  his  farm.?  Ans.  30  acres. 

13.  A  company  of  15  persons  engaged  a  dinner  at  a  hotel,  but 
before  paying  the  bill  5  of  the  company  withdrew  by  which  each 
person's  bill  was  augmented  $-J;  what  was  the  bill?        Ans.  $15. 

14.  A  man  sold  his  horse  and  sleigh  for  $200,  and  f  of  this  is 
8  times  what  his  sleigh  cost,  and  the  horse  cost  10  times  as  much 
as  the  sleigh  ;  required  the  cost  of  each. 

Ans.  horse,  $200;  sleigh,  $20, 

15.  A  went  to  a  store  and  borrowed  as  much  as   he   had,  and 
spent  4  cents;  he  then  went  to  another   store   and   did   the  same, 
and  then  had  4  cents  remaining;   how   much   money   had   he  at 
first?  Ans.  4  cents. 

16.  A  lady  being  asked  her  age,  said  that  if  her  age  were  in- 
creased by  its  ^,  the  sum  would  equal  3  times   her  age   12  years 
ago;  what  was  her  age?  Ans.  20. 

17.  A  lady  being  asked  the  hour  of  day,  replied  that  f  of  the 
time  past  noon  equaled  |-  of  the  time  to  midnight,  minus  -J  of  an 
hour;  what  was  the  time?  Ans.  6  o'clock,  P.  M. 


PROBLEMS.  179 

18.  What  is  the  hour  of  day  if  -j-  of  the   time   to   noon  equals 
the  time  past  midnight?  Ans.  9  o'clock,  A.  M. 

19.  A  person  being  asked  the  time  of  day,  said  f  of  the  time 
to  midnight  equals  the  time  past  midnight  ;    what  was  the  time? 

Ans.  9  o'clock,  A.  M. 

20.  A  traveler  on  a  train  notices  that  4J  times  the  number  of 
spaces  between  the  telegraph  poles  that  he  passes  in  a   minute  is 
the  rate  of  the  train  in  miles  per   hour.     How    far   are   the  poles 
apart?  Ans.  198  feet. 

21.  C's  age  at  A's  birth  was  5-J  times  B's  age,  and  now  is  the 
sum  of  A's  and  B's  ages,  but  if  A  were  now  3  years  younger  and 
B  4  years  older,  A's  age  would  be  f  of  B's  age.    Find  their  ages. 

Ans.  A's,  72  years;  B's,  88  years;  C's,  160  years. 

22.  In  the  above  problem  change  the  last  and  to  or,  and  what 
are  their  ages?  Ans.  A's,  36  ;  B's,  44,  and  C's,  80. 

23.  I  have  four  casks,  A,   B,    C,  and   D   respectively.     Find 
the  capacity  of  each,  if  f  of  A  fills  B,  f  of  B  fills   C,    and   €  fills 
T9g-  of  D;  but  A  will  fill  C  and  D  and  15  quarts  remaining. 

Ans.  A  35  gal.,  B  15,  C  11^,  and  D  20. 


24.  A  man  and  a  boy  can  do  a   certain   work   in   20  days  :  if 
the  boy  rests  5^  days  it  will  take  them   22^-  days;   in    what  time 
can  each  do  it?  Ans.  The  man,  36  da.  ;  the  boy,  45  da. 

25.  A  can  do  a  job  of  work  in  40  days,   B   in   60  days;  after 
both  work  3  days,  A  leaves  ;  when  must  he  return  that  the  work 
may  occupy  but  30  days?  Ans.  10  days. 

26.  If  8  men  or  15  boys  plow  a   field   in   15   days   of  9-J  hr., 
how  many  boys  must  assist  16  men  to  do  the  work  in   5   days  of 
10  hr.  each?    '  Ans.  12  boys. 

27.  Bought  10  bu.  of  potatoes  and  20  bu.    of  apples   for  $11  ; 
at  another  time  20  bu.  of  potatoes  and  10  bu.  of  apples    for  $13  ; 
what  did  I  pay  for  each  per  bu.  ? 

Ans.  Apples  30/,  potatoes  50/. 

28.  A  farmer  sold  17  bu.  of  barley  and    13   bu.   of  wheat  for 
$31.55,  getting   35/  a  bu.    more   for   wheat   than    for   the  barley. 
Find  the  price  of  each  per  bu. 

Ans.     Barley  90/,  wheat  $1.25. 

29.  After  losing  f  of  my  money  I  earned  $12;  I  then  spent  f 
of  what  I  had  and  found  I  had  $36   less   than   I   lost;   ho\v  much 
money  had  I  at  first?  Ans.  $60. 

30.  In  a  company  of  87,  the  children  are  -J  of  the  women,  and 
the  women  f  of  the  men;  how  many  are  there  of  each? 

Ans.  54  men,  24  women,  and  9  children. 


180  FINKEL'S   SOLUTION  BOOK. 

31.  If  4  horses  or  6  cows  can  be  kept  10  days  on  a  ton  of  hay, 
how  long  will  it  last  2  horses  and  12  cows?  Ans.  4  days, 

32.  A,  B,  and  C  buy  4  loaves  of  bread,  A  paying   5   cents,  B 
8  cen^s,  and  C  11  cents.     They  eat  3  loaves   and    sell   the  fourth 
to  D  for  24  cents.     Divide  the  24  cents  equitably. 

Ans.     A  5  cents,  B  8  cents,  and  C  11  cents. 

33.  A  and  B  are  at  opposite  points  of  a  field  135  rods  in  com- 
pass, and  start  to  go  around  in  the  same  direction,  A  at   the  rate 
of  11  rods  in  2  minutes   and    B    17   rods   in   3   minutes.     In  how 
many  rounds  will  one  overtake  the  other?         Ans.  B  17  rounds. 

34.  If  a  piece  of  work  can  be  finished  in  45   days  by   35  men 
and  the  men  drop  off  7  at  a  time  every  15  days,  how  long  will  it 
be  before  the  work  is  completed?  Ans.  75  days. 

35  A  watch  which  loses  5  min  a  day  was  set  right  at  12  M., 
July  24th.  What  will  be  the  true  time  on  the  30th,  when  the 
hands  of  that  watch  point  to  12?  Ans.  12:30-^  P.  M. 

36.  A  seed  is  planted.     Suppose  at  the  end  of  3  years   it  pro- 
duces'a  seed,  and  on  each  year  thereafter  each  of  which  when  3 
years  old   produce  a   seed   yearly.     All   the   seeds   produced:  do 
likewise  ;  how  many  seeds  will  be  produced  in  21  years? 

Ans.  1872. 

37.  The  circumference  of  a  circle  is  390  rods.     A,   B,  and  C 
start  to  go  around  at  the  same  time.    A  walks  7  rods  per  minute, 
B  13  rods  per  minute  in  the  same  direction  ;  C  walks  19  rods  per 
minute  in   the   opposite   direction.     In   how   many   minutes  will 
they  meet?  Ans.  195  min. 

38.  If  12  men  can  empty  a  cistern   into   which   water   is  run- 
ning at  a  uniform  rate,  in  40  min.,  and  15  men  can  empty  it  in  30 
min.,  how  long  will  it  require  18  men  .to  empty  it? 

Ans.  24  min. 

39.  Four  men  A,  B,  C,  and  D,  agree  to  do  a  piece  of  work  in 
130  days.     A  gets  42d.,  B  45d.,  C  48d.,   and   D   5id.,   for  every 
day  they  worked,  and  when   they   were   paid   each    man   has  the 
same  amount.     How  many  days  did  each  work?  [da. 
Ans.     A  35ffff  da.,  B  33f \\\  da,  C  31^H  da->  and   D  29fftl 

40.  A  fountain  has  four  receiving  pipes,  A,  B,  C,  and  D;   A, 
B,  and  C  will  fill  it  in  6  hours;  B,  C,  and   D   in   8   hours;   C,  D, 
and  A  in  10  hours;  and  D,  A,  and  B  in  12  hr.:  it  also  has  four  dis- 
charging pipes,  W,  X,  Y,  and  Z  ;  W,  X,  and  Y  will  empty   it  in 
6  hours;  X,  Y,  Z  in  5  hours;  Y,  Z,  and   W   in   4   hours  ;   and  Z, 
W,  and  X   in   3   hours.     Suppose   the   pipes   all   open,   and   the 
fountain  full,  in  what  time  will  it  be  emptied?     Ans.  6TV  hours. 


ALLIGATION.  181 


CHAPTER  XVIII. 

ALLIGATION. 

1.  Alligation  is  the  process   employed   in   the  solution  of 
problems   relating   to   the   compounding   of  articles  of  different 
values  or  qualities. 

/I77.       .  .         (  1.  Alligation  Medial. 

2.  All^gat^on  \  2   Alli|ation  Alternate. 

I.  ALLIGATION  MEDIAL. 

1.  Alligation  Medial  is  the  process  of  finding  the  mean, 
or  average,  rate  of  a  mixture  composed  of  articles  of  different 
values  or  qualities,  the  quantity  and  rate  of  each  being  given. 

I.     A  grocer  mixed  120  lb.  of  sugar  at  5/  a  pound,  150  lb.  at 
6/.,  and  130  lb.  at  10/.;  what  is  the  value  of  a  pound  of 

the  mixture? 

120  lb.  @5/=$6.00, 
150  lb.  @6/=$9.00,  and 
130  lb. 


II 


1. 
2. 
3. 
4.  400  lb.  is  worth  $28.00. 


III. 


5.  .-.  1  lb.  is  worth  $28-r-400=$.07=7  cents. 
.'.  One  pound  of  the  mixture  is  worth  7  cents. 

,  (  Stod.  Comp.  A.,  p.  244,  p™b.  3.  ) 

II.  ALLIGATION  ALTERNATE. 

1.  Alligation  Alternate  is  the  process  of  finding  in 
what  ratio,  one  to  another,  articles  of  different  rates  of  quality  or 
value  must  be  taken  to  compose  a  mixture  of  a  given  mean,  or 
average,  rate  of  quality  or  value. 

CASE  I. 

Given  the  value  of  several  ingredients,  to  make  a  compound  of 
a  given  value. 

I.  What  relative  quantities  of  tea,  worth  25,  27,  30,  32,  and 
45  cents  per  lb.  must  be  taken  for  a  mixture  worth  28 
cents  per  lb. 


Dif. 


Bal. 


SOLUTION.  —  In    average, 
the    principle    is,     that    the 
gains    and   loses   are   equal.        ' 
We  write  the  average  price 
and  the  particular  values  25, 

'25X 
27/ 
30/ 

32X 

45X 

3X 

I/ 

2/ 
4X 
17/ 

2  lb. 
3  lb. 

17  lb. 

31b. 

i 

4  lb. 
1  lb. 

19  lb. 
4  lb. 
31b. 
1  lb. 
3  lb. 

27,  30,  32,  and  45  as  in  the  margin.     This   is   only   a  convenient 


182  FINKEL'S   SOLUTION   BOOK. 

arrangement  of  the  operation.  Now  one  pound  bought  for  25/ 
and  sold  in  a  mixture  worth  28/  there  is  a  gain  of  28/ — 25/,  or 
3/;  one  pound  bought  at  27/  and  sold  in  a  mixture  worth  28/,, 
there  is  a  gain  of  28/ — 27/,  or  I/;  one  pound  bought  at  30/  and 
sold  in  a  mixture  worth  28/  there  is  a  loss  of  30/ — 28/,  or  2/ ; 
one  pound  bought  at  32/  and  sold  in  a  mixture  worth  28/,  there 
is  a  loss  of  32/ — 28/,  or  4/;  and  one  pound  bought  at  45/  and 
sold  in  a  mixture  worth  28/  there  is  a  loss  of  45/ — 28/,  or  17/V 
Since  the  gains  and  losses  are  equal,  we  must  take  the  ingredi- 
ents composing  this  mixture  in  such  a  proportion  as  to  make  the 
gains  and  losses  balance.  We  will  first  balance  the  25/  tea  and 
the  30,£  tea.  Since  we  gain  3/  a  pound  on  the  25/  tea,  and  lose 
2/  on  the  30/  tea,  how  many  pounds  of  each  must  we  take  so 
that  the  gain  and  loss  on  these  two  kinds  may  be  equal?  Evi- 
dently, we  should  gain  6/  and  lose  6/.  To  find  this,  we  simply 
find  the  L.  C.  M.  of  3  and  2.  Now  if  we  gain  3/  on  one  pound 
of  the  25/  tea,  to  gain  6/,  we  must  take  as  many  pounds  as  3/ 
is  contained  in  6/,  which  are  2  Ib.  If  we  lose  2/  on  one  pound 
of  the  30/  tea,  to  lose  6^,  we  must  take  as  many  pounds  as  2/ is 
contained  in  6/,  which  are  3  Ib.  Next,  balance  the  25-cent  tea 
and  the  45-cent  tea.  The  L.  C.  M.  of  3/  and  17/  is  51/.  Now 
if  we  gain  3/  on  one  pound  of  the  25-cent  ^ea  to  gain  51/,  we 
must  take  as  many  pounds  as  3/  is  contained  in  51/  which  are 
17  Ib.  If  we  lose  17/  on  one  pound  of  the  45-cent  tea,  to  lose 
51/,  we  must  take  as  many  pounds  as  17/  is  contained  in  51/ 
which  are  3  Ib.  Next,  balance  the  27-cent  tea  and  the  32-cent 
tea.  The  L.  C.  M.  of  1^  and  4/  is  4/.  If  we  gain  I/  on  one 
pound  of  the  27-cent  tea,  to  gain  4/,  we  must  take  as  many 
pounds  as  I/  is  contained  in  4/,  which  are  4  Ib.  If  we  lose  4/ 
on  one  pound  of  the  32-cenf  tea,  if  balances  the  gain  on  the  27- 
cent  tea.  Placing  the  number  of  pounds  to  be  taken  of  each, 
kind  as  shown  above,  and  then  adding  horizontally,  we  have  19 
Ib.  at  25/,  4  Ib.  at  27/,  3  Ib.  at  30/,  1  Ib.  at  32/,  and  3  Ib.  at 
45/.  It  is  not  necessary  to  balance  them  in  any  particular  order. 
All  that  must  be  observed,  is  that  all  the  ingredients  be  used  in 
balancing. 

Note. — To  prove  the  problem,  use  Alligation  Medial. 


CASE  II. 

To  proportionate  the  parts,  one  or  more  of  the  quantities,  but 
not  the  amount  of  the  combination,  being  given. 

I.  How  many  bushels  of  hops,  worth  respectively  50,  60, 
and  75/  per  bushel,  with  100  bushels  at  40/  per  bushel,  will 
make  a  mixture  worth  65^"  a  bushel? 


ALLIGATION. 


183 


Dif. 


40/I25/ 


65/. 


50/ 
60X 
75> 


15/ 

r>p 

W? 


Bal. 


2  bu. 

2  bu. 

2  bu. 

2bu. 

2  bu. 

2  bu. 

5  bu. 

3  bu. 

1  bu 

9  bu. 

-100  bu. 
100  bu. 
100  bu. 
.450  bu. 


Dif.        {       Bal. 


B.     65/. 


40/ 
50/ 
60/ 

25/ 
15/ 

2bu. 

2  bu. 
3bu. 

10/ 

bu. 


1100  bu. 
2bu. 
2bu. 
254  bu. 


SOLUTION. — In  this  solution,  we  proceed  as  in  Case  I.  In  A,. 
we  obtain  the  relative  amounts  to  be  used  of  each  kind,  which  is 
2  bu.  at  40/,  2  bu.  at  50/,  2  bu.  at  60/,  and  9  bu.  at  75/.  But 
we  are  to  have  100  bu.  of  the  first  kind.  Hence,  we  must  multi- 
ply these  results  by  100-i-2,  or  50.  Doing  this,  we  obtain  100 
bu.  at  40/,  100  bu.  at  50/,  100  bu.  at  60/,  and  450  bu.  at  75/. 

Since  either  or  both  of  the  balancing  columns,  except  the  first, 
may  be  multiplied  by  any  number  whatever  without  affecting 
the  average,  it  follows  that  there  are  an  infinite  number  of  re- 
sults satisfying  the  conditions  of  the  problem.  Since  we  are  to 
have  100  bu.  at  40/,  the  first  column  can  be  multiplied  by 
only  50. 

In  B,  we  have  multiplied  the  first  column  by  50  and  added  in 
the  results  in  the  other  two  columns.  This  gives  us  100  bu.  at 
40/,  2  bu.  at  50/,  2  bu.  at  60/,  and  254  bu.  at  75/.  The  second 
and  third  columns  may  be  multiplied  by  any  number  whatever. 
But  the  first  must  always  must  be  multiplied  by  50,  because  we 
are  to  have  100  bu.  at  40  cents  per  bushel. 

(R.  H.  A., p.  338,prob.  2.) 

I.  How  much  lead,  specific  gravity  11,  with  ^  oz.  copper, 
sp.  gr.  9,  can  be  put  on  12  oz.  of  cork,  sp.  gr.  ^,  so  that  the  three 
will  just  float,  that  is,  have  a  sp.  gr.  (1)  the  same  as  water? 


r 

3 

3 

x| 

8 

1  0 

9 

Ti 

I                  J 

oz.=2  Ib.  1    oz. 


12  oz. 


SOLUTION.— The    specific    gravity    of  any    body    is   the  ratio 
which  shows  how  many  times  heavier  the  body  is  than   an  equal 


184  FINKEL'S   SOLUTION  BOOK. 

volume  of  water.  Thus,  when  we  say  that  the  sp^inY  gravity 
of  lead  is  11,  we  mean  that  a  cubic  inch,  a  cubic  foot,  a  cubic 
yard,  or  any  quantity  whatever  is  11  times  as  heavy  as  an  equal 
quantity  of  water. 

Now  if  a  cubic  inch  (say)  of  lead  be  immersed  in  water,  it 
will  displace  a  cubic  inch  of  water ;  and  since  it  weighs  11  times 
as  much  as  a  cubic  inch  of  water,  it  displaces  Jy  of  its  own 
weight.  Hence,  to  have  equal  weights  of  water  and  lead  we 
must  take  only  Jy  as  much  lead  as  water.  Now  since  a  volume 
of  water  and  y^  as  much  lead  have  the  same  weight,  and  in 
the  proper  combination  have  a  volume  of  1,  since  the  sp.  gr.  oi 
the  combination  is  1,  there  is  a  loss  of  1 — Jy,  or  T^,  in  volume  on 
the  part  of  the  lead.  For  the  same  reason,  there  is  a  loss  of  f  in 
volume  on  the  part  of  the  copper,  and  3  on  the  part  of  the  cork. 
Balancing,  we  see  that  we  must  take  3  volumes  of  lead  with  T§- 
volumes  of  cork,  a  unit  volume  of  water  being  the  basis,  in  order 
that  the  two  substances  will  just  float,  /.  e.,  have  a  specific  grav- 
ity (1).  In  like  manner,  we  must  take  3  volumes  of  copper 
with  -|  volumes  of  cork.  Now  since  we  must  always  take  3  vol- 
umes of  lead  for  every  T-2-  volumes  of  cork,  it  is  evident  that  the 
weights  of  the  substances  are  in  the  same  proportion.  Hence, 
we  may  say,  we  must  take  3  oz.  of  lead  with  every  y-f-  oz.  of 
cork,  and  3  oz.  of  copper  with  every  -f  oz.  of  cork. 

But  we  are  to  have  only  -J-  oz.  of  copper.  Hence,  we  must 
multiply  the  second  balancing  column  by  some  number  that  will 
give  us  -J  oz.  of  copper,  /.  £.,  we  must  multiply  3  by  some  number 
that  will  give  us  -J.  The  number  by  which  we  must  multiply  is 
-|~j-3— -g-.  But  multiplying  -|  by  -^,  we  get  ^4T  oz.  of  cork.  But 
we  are  to  have  altogether  12  oz.  of  cork.  Hence  we  must  yet 
have  12  oz. — 24T  oz.=  3^T°  oz.  To  produce  this,  we  must  multi- 
ply T^  by  some  number  that  will  give  3f2T°  oz.  This  number  is 
S^O^H^SSJJ.  But  we  must  also  multiply  3  by  $fj.  This  will 
.give  us  39^  oz.=2  lb.  7-§-  oz.  of  lead.  Hence,  we  must  use  2  Ib.  7-j- 
•oz.  of  lead,  so  that  the  three  will  just  float. 

(jR.  H.  A., p.  339,prob.  7.) 

I.  How  many  shares  of  stock,  at  40%,  must  A  buy,  who  has 
bought  120  shares,  at  74%,  150  shares,  at  68%,  and  130 
shares,  at  54%,  so  that  he  may  sell  the  whole  at  60%, 
and  gain  20%  ? 


>•• 

'(1.)   100  %=the  average  cost. 
(2.)  20%=gain. 
(3.)   120%—  the  average  selling  price. 
(4.)  60%=the  average  selling  price. 
(5.)  /   120%=60%. 
(6.)   1^=^  of  60%=^%. 
(7.)   100%=100  times  ^%=50%,  the  average 

cost. 

ALLIGATION. 


185' 


II. 


120  shares  @   74%== 

150  shares  @  68%=10200%. 

130  shares  @  54%=  7020%. 
.-.  400  shares  are  worth  26100%,  and 
1  share  is  worth  26100%-r-400=65i% ,  the  average. 


3.     50% 


40  % 
65i% 


10  % 


shares. 


10  shares. 


X40=< 


-610  shares. 
.400  shares. 


III.     .-.  He  must  take  610  shares.       (R.  H.  A.,  p.  339,  prob.  8.) 


Explanation. — Since  60%  is  the  average  selling  price,  and  his  gain  is 
it  is  evident  that  his  average  cost  is  60%-M.20,  or  50%.  In  step  3,  we  find 
that  the  average  cost  of  the  400  shares  is  65^%.  Hence,  the  problem  is  the 
same  as  to  find  how  many  shares  at  40%,  must  A  buy  who  has  400  shares  at 
at  an  average  of  65^%  so  that  his  average  cost  will  be  50%.  Balancing, 
we  find  that  he  must  take  15£  shares  at  40%  with  10  shares  at  65 \%.  But 
he  has  400  shares  at  65^%.  Hence,  we  must  multiply  the  balancing  col- 
umn by  400-MO,  or  40.  This  gives  610  shares  at  40%. 


CASE  III. 

To  proportion  the  parts,  the  amount  of  the  whole  combination 
being  given. 


II. 


III. 


How  many  barrels  of  flour,  at  $8,  and  $8.50,  with  300 
bbl.  at  $7.50,  800  bbl.  at  $7.80,  and  400  bbl.  at  $7.65, 
will  make  2000  bbl.  at  $7.85  a  bbl.  ? 


300  bbl. 
800  bbl. 
400  bbl. 


$7.50  a  bbl.=$2250. 
$7.80  a  bbl.=$6240. 
$7.65  a  bbl.=$3060. 




4.  ,'.  1500 *bbl.  are  worth  $11550. 

5.  $7.85=the  average  price  per  bbl.  of  2000  bbl. 

6.  /.  $15700=2000  X$7.85=the  value  of  2000  bbl. 

7.  .'.  $15700— $11550=$4150=the  value  of  2000  bbl.— 1500 

bbl.,  or  500  bbl. 
8.30=$4150-f-500=the  average  value  of  1  bbl. 


8.  . 

9.  $8.30 


$8.00 


$.302  bbl.l 


X(  500^-5  )= 


200  bbl. 
300  bbl. 


$8.50  $.203  bbl.| 
5  bbl. 

1.  200  bbl.  at  $8.00  per  bbl.  must  be  taken  with 

2.  300  bbl.  at  $8.50  per  bbl. 

(R.  H.  A.,  p.  339,prob. 


I.  A  dealer  in  stock  can  buy  100  animals  for  $400,  at  the  fol- 
lowing rates:  calves,  $9;  hogs,  $2;  lambs,  $1;  how  many  may 
he  take  of  each  kind  ? 


186 


FINKEL'S  SOLUTION  BOOK. 


BaL 


$1 

$3i5  lambs. 

3 

10(17 

24 

31 

38 

45 

62 

59 

$4 

$2 

<feoi 
$z, 

5  hogs. 

68 

60 

52 

44 

36 

28 

20 

12 

4 

$9 

$5:3  calves. 

2  calves. 

29 

30 

31 

32 

33 

34 

35 

36 

37 

8 

Explanation. — A  lamb  bought  for  $1  and  sold  for  $4  is  a  gain  of  $3;  a 
hog  bought  for  $2  and  sold  for  $4  is  a  gain  of  $2;  and  a  calf  bought  for  $9 
and  sold  for  $4  is  a  loss  of  $5.  We  must  make  the  gains  and  loses  equal. 
The  L.  C.  M.  of  $3  and  $5  is  $15.  If  we  gain  $3  on  one  lamb  to  gain  $15 
we  must  take  as  many  lambs  as  $3  is  contained  in  $15,  which  are  5  Iambs. 
If  we  lose  $5  on  one  calf,  to  lose  $15,  we  must  take  as  many  calves  as  $5  is 
contained  in  $15,  which  are  3  calves.  The  L.  C.  M.  of  $2  and  $5  is  $10.  If 
we  gain  $2  on  one  hog,  to  gain  $10,  we  must  take  as  many  hogs  as  $2  is 
contained  in  $10,  which  are  5  hogs.  If  we  lose  $5  on  one  calf,  to  lose  $10, 
we  must  take  as  many  calves  as  $5  is  contained  in  $10,  which  are  2  calves. 
Adding  the  balancing  columns,  considering  them  as  abstract  numbers,  we 
have  8  and  7.  8+7=15.  100-i-15=6f.  .'•  Multiplying  each  balancing 
column  by  6|,  will  give  33£  lambs,  33£  hogs,  and  33£  calves.  But  this  result 
is  not  compatible  with  the  nature  of  the  problem.  Hence  we  must  see  if 
we  can  take  a  number  of  8's  and  a  number  of  7's  that  will  make  100.  By 
trial,  we  find  that  two  8's  and  twelve  7's  will  make  100.  Hence,  multiplying  the 
first  column  by  2  and  the  second  by  12,  and  adding  the  columns  horizon- 
tally, we  have  for  our  result,  10  lambs,  60  hogs,  and  30  calves.  Again,  we 
•find,  by  trying  three  8's,  four  8's,  and  so  on,  that  nine  8's  taken  from  100, 
will  leave  28  which  is  four  7's.  Hence,  nine  8's  and  four  7's  will  make  100. 
Then,  multiplying  the  first  column  by  9  and  the  second  by  4,  and  adding 
the  columns  horizontally,  we  have  for  a  second  result  45  lambs,  20  hogs, 
and  35  calves.  Now  these  are  the  only  answers  that  can  be  obtained  by 
taking  an  integral  number  of  8's  and  integral  number  of  7's  to  make  100. 
But  other  answers  may  be  obtained  by  taking  8  a  fractional  number  of 
times,  and  7  a  fractional  number  of  times  to  make  100.  Suppose,  for  illus- 
tration, we  try  to  take  a  number  of  thirds  8  times.  We  find  that  8  taken  6- 
third  times  and  7  taken  36  third  times  will  make  100.  Multiplying  the 
first  column  by  f  and  the  second  by  3^,  and  adding  the  columns  horizon- 
tally, we  have,  for  a  result,  10  lambs,  60  hogs,  and  30  calves — the  same  as 
that  obtained  by  taking  8  twice  and  7  twelve  times.  Again,  we  find,  that 
8  taken  13  third  times  and  7  taken  28-third  times  will  make  100.  Multiply- 
ing and  adding  as  before  we  find  that  our  results  are  fractional.  Hence,  we 
can  not  take  a  fraction  whose  denominator  is  three.  It  is  clear  that  we 
must  take  a  fraction  whose  denominator  will  reduce  to  unity,  when  being 
multiplied  by  5.  Hence,  if  we  try  to  take  8  a  number  of  fifths  times  and  7 
a  number  of  fifths  times  to  make  100,  our  results  will  all  be  integral.  -By 
trial,  we  find  that  8  taken  3-fifths  times  and  7  taken  68-fifths  times  will 
make  100.  Multiplying  and  adding  as  before,  we  have,  for  our  results,  3 
lambs,  68  hogs,  and  29  calves.  Again,  we  find  that  8  taken  10- fifths  times 
and  7  taken  60-fifths  times,  will  make  100.  Multiplying  and  adding  as  be- 
fore, we  have,  for  results,  10  lambs,  60  hogs,  and  30  calves.  Again,  by  trial, 
we  find  that  8  taken  17-fifths  and  7  taken  52-fifths  times  will  make  100. 
Multiplying  the  first  column  by  y  and  the  second  by  ^,  and  adding  the  col- 
umns horizontally,  we  have,  for  results,  17  lambs,  52  hogs,  and  31  calves. 
Continuing  the  process,  we  find  nine  admissible  answers.  These  are 
the  only  answers,  satisfying  the  nature  of  the  problem. 


SYSTEMS  OF  NOTATION. 


187 


CHAPTER   XIX. 

SYSTEMS  OF  NOTATION. 


1.     A   System  of  Notation  is    a  method    of  expressing 

numbers  by  means  of  a  series  of  powers  of  some   fixed   number 

called    the  Radix,  or  Base  of  the  scale  in  which    the   different 
numbers   are  expressed. 


2.     The  Jtaclioc  of  any  system  is  the  number  of  units  of  one 
order  which  makes  one  of  the  next  higher. 


3. 


Names  of  Systems. 

Radix. 

Names  of  Systems. 

Radix. 

Unitary 

1 

Nonary 

9 

Binary          -      .   - 

2 

Decimal,  or  Denary 

10 

Ternary 

3 

Undenary 

11 

Quaternary 

4 

Duodenary, 

12 

Quinary 

5 

Vigesimal, 

20 

Senary 

6 

Trigesimal, 

30 

Septenary 

7 

Sexagesimal, 

60 

Octonary     - 

8 

Centesimal, 

100 

4.  In  writing  any  number   in   a   uniform   scale,  as   many  dis- 
tinct characters,  or  symbols,  are  required  as  there  are  units  in  the 
radix  of  the  given  system.     Thus,  in  the  decimal  system,  10  char- 
acters  are   required;  in   the  ternary,  3;   viz.,  1,  2,  and   0;  in  the 
senary,  6;  viz.,  1,  2,  3,  4,  5,  and  0;  and  so  on. 

5.  Let  r  be   the   radix   of  any  system,  then   any  number,  N, 
may  be  expressed  in  the  form, 


the  co-efficients  £,£,£, 


2  -f-  qr-\-  s,  in  which 
are  each  less  than  r. 


To   express    an  integral   number  in  a  proposed  scale  :     Divide 
the  number  by  the  radix,  then  the  quotient  by  the  radix,  and  so 
the  successive  remainders  taken  in  order  will  be  the  successive 


on  • 


digits  beginning  from  units  place. 


188  FINKEL'S   SOLUTION   BOOK. 

I.     Express  the  common  number,  75432,  in  the  senary  system. 

1.  6)75432 

2.  6)12572+0 


II. 


3.  6)2095+2 

4.  6)349+1 


5.  6)58+1 

6.  6)9+4 
L7.  1+3 

III.     .-.  75432  in  the  decimal  system=1341120  expressed   in  the 

senary  system. 

I.     Transform  3256  from  a  scale  whose  radix  is   7,   to   a  scale 
whose  radix  is  12. 

1.  12)3256 

2.  12)166+4 

3.  12)11+1 


II 


4.  0+8 

III.  .-.  3256  in  the  septenary  system=814  in  the  duodenary 
system. 

Explanation. — In  the  senary  system,  7  units  of  one  order  make  one  of  the 
next  higher.  Hence,  3  units  of  the  fourth  order  =  7X3,  or  21,  units  of  the 
third  order.  21  units +2  units  =23  units.  23-7-12  =  1,  with  a  remainder 
11.  11  units  of  the  third  order  =77  units  of  the  second  order.  77  units 
+5  units=82  unrts.  82+12=6,  with  a  remainder  10.  10  units  of  the 
second  order=70  units  of  the  first  order.  70  units  +6  units=76  units.  76 
+12= 6,  with  a  remainder  4.  Hence,  the  first  quotient  is  166,  with  a  re- 
mainder 4.  Treat  this  quotient  in  like  manner,  and  so  on,  until  a  quo- 
tient is  obtained,  that  is  less  than  12. 

I.  What  is  the  sum  of  45324502  and  25405534,  in  the  senary 
system  ? 

45324502 

25405534 

115134440 

Explanation. — 4+2=6.  6-^-6=1,  with  no  remainder.  Write  the  0  and 
carry  the  1.  3+1=4.  Write  the  4.  5+5=10.  10+6=1,  with  a  remainder 
4.  Write  the  4  and  carry  the  1.  5+4+1=10.  10+6=1,  with  a  remain- 
der 4.  Write  the  4  and  carry  the  1.  0+2+1=3.  Write  the  3.  4+3=7. 
7+6+1  with  a  remainder  1.  Write  1  and  carry  1  5+5+1=11.  ll-r-6= 
1,  with  a  remainder  5.  Write  the  5  and  carry  the  1.  2+4+1=7.  7+6= 
1,  with  a  remainder  1.  Write  1  and  carry  1.  The  result  is  115134440. 

I.  What  is  the  difference  between  24502  and  5534  in  the 
octonary  system? 

24502 

5534 

16746 

Explanation. — 4  cannot  be  taken  from  2.  Hence,  borrow  one  unit  from 
a  higher  denomination.  Then  (2+8)— 6=4.  (8—1)— 3=4.  5  from  (4+8) 
—7.  5  from  (3+8)=6.  Hence,  the  result  is  16746. 


SYSTEMS  OF  NOTATION.  189 

I.     Transform  3413  from  the  scale  of  6  to  the  scale  of  7. 

1.  7)3413 
II.<[2.     7)310+3 
3.      7)24+3 
2+2 

III.     .-.  3413  in  the  senary  system=2233  in  the   septenary  sys- 
tem. 

I.     Multiply  24305  by  34120  in  the  senary  system. 

24305 

34120 
530140 
24305 
150032 
121323 
1411103040 

Explanation, — Multiplying  5  by  2  gives  10.  HK-6— 1,  with  a  remainder 
4.  Write  4  and  carry  1  to  the  next  order.  2  times  0—0.  0+1—1.  Write 
the  1.  2  times  3=6.  6-4-6=1,  with  a  remainder  0.  Write  the  0  and 
carry  the  1  to  the  next  higher  order.  2  times  4=8.  84-1—9.  9-*-6=l, 
with  a  remainder  3.  Write  3  and  carry  the  1  to  the  next  higher  order. 
2  times  2=4.  4+1=5.  Write  5.  Multiply  in  like  manner  bv  1,  4,  and  3. 
Add  the  partial  products,  remembering  that  6  units  of  one  order,  in  the  sen- 
ary system,  uniformly  make  one  of  the  next  higher. 

I.     Multiply  2483  by  589  in  the   undenary   system,   or  system 
whose  radix  is  11. 

We  must  represent  10  by  some  character.     Let  it  be  t. 

2483 
589 


1*985 
1/502 
11184 
13322/5 

Explanation. — In  the  undenary  system,  11  units  of  one  order  uniformly 
make  one  of  the  next  higher  order.  9  times  3=27.  27 -f- 11=2,  with  a  re- 
mainder 5.  Write  5  and  carry  the  2  to  the  next  higher  order,  or  second  or- 
der. 9  times  8=72.  72+2=74.  74^-11=6,  with  a  remainder  8.  Write  8 
and  carry  the  6  to  the  next  higher  order,  or  third  order.  9  times  4=36. 
36+6=42.  42-5-11=3,  with  a  remainder  9.  Write  9  and  carry  the  3  to  the 
next  higher  order,  or  the  fourth  order.  9  times  2=18.  18+3=21.  21-j-ll 
=1,  with  a  remainder  /.  Write  t  and  carry  the  1  to  the  next  higher  order. 
Multiply  in  like  manner  by  8  and  5.  Add  the  partial  products,  remember- 
ing that  11  units  of  one  order  equals  one  of  the  next  higher.  Wherever 
10  occurs,  it  must  be  represented  by  a  single  character  t. 


190  FINKEL'S   SOLUTION  BOOK. 

I.     Divide  1184323  by  589  in  the  duodenary  system. 

In  the  duodenary  system,  we  must  have  12  characters  ;   viz.,  1, 
2,  3,  4,  5,  6,  7,  8,  9,  t,  e,  and  0.     /  represents  10  and  e,  11. 
589)1184323(2486 


_ 
22/3 


3*32 
39/0 

1523 

1523 

Explanation.  —  In  the  duodenary  system,  12  units  of  one  order  make  one 
of  the  next  higher.  1184  will  'contain  589,  2  times.  Then  multiply  the 
divisor,  589,  by  2  thus:  2  times  9=18.  18-4-12=1,  with  a  remainder  6. 
Write  the  6  and  carry  the  1.  2  times  8=16.  16+1=17.  '  17-*-  12=1,  with  a 
remainder  5.  Write  the  5  and  carry  the  1.  2  times  5=/.  t-}-l=e. 
Write  the  e.  Then  subtract.  6  from  (12+4)=^,  5  from  7=2,  and  e  from  (12 


Hence,  the  first  partial  dividend  is  22  /.  Bring  down  3.  Then  22t3  will 
contain  589,  4  times.  Multiply  as  before.  By  continuing  the  operation  we 
obtain  2483  for  a  quotient. 

I.     Divide  95088918  by  #4,  in  the  duodenary  system. 


I.     Extract  the  square  root  of  11122441  in  the  senary  system. 


11122441(2405 
2X2=_4 
44  ]  312 
304 


hi 

Li 


2X24=52  0 
2X240=520  5 


42441 
42441 


Explanation. — The  greatest  square  in  11  expressed  in  the  senary  system 
is  4.  Subtracting  and  bringing  down  the  next  period,  we  have  312  for  the 
next  partial  dividend.  Doubling  the  root  already  found  and  finding  how 
many  times  it  is  contained  in  312  expressed  in  the  senary  system,  we  find  it 
is  4.  Continuing  the  process  the  same  as  in  the  decimal  system,  the  re- 
sult is  2405. 


SYSTEMS  OF  NOTATION. 


191 


I.     Extract  the  square  root  of  11000000100001   in  the  binary 
system. 


I. 


II. 


III. 


I. 


III. 


11000000100001(1101111 

1 


101  H 

300 
101 

11001 
110101 
1101101 
11011101 

110000 
11001 

1011110 
110101 

10100100 
1101101 

11011101 
11011101 

(  Todhunter's  Algebra,  p.  855,  Ex.  28.) 
Find  in  what  scale,  or  system,  95  is  denoted  by  137. 

1.  Let  r=the  radix  of  the  system.     Then 
2. 


3.  r2+3r=95—  7=88,  and 

4.  r24-3r4-f=88+f=3£  1,  by  completing  the  square. 

5.  7"-f-f=V>  by  extracting  the  square  root,  and 

6.  r±=y—  |=y=8,  the  radix  of  the  system. 

/.  95  is  denoted  by  137  in  the  octonary  system. 

(  Todhunter's  Alg.,  p.  255,  prob. 

Find  in  what  system  1331  is  denoted  by  1000. 

1.  Let  r=the  radix  of  the  system.     Then 

2.  r 

3.  r4=1331.     Whence 

4.  r= 


=11,  the  radix  of  the  system. 

/.  1331  is  denoted  by  1000  in  the  undenary  system. 

(  Todhunter's  Alg.,  p.  255,  prob.  28.) 


192 


FINKEL'S   SOLUTION  BOOK. 


CHAPTER  XX. 

MENSURATION. 

1,  Mensuration  is  that   branch    of  applied    mathematics 
which  treats  of  geometrical  magnitudes. 

2.  Geometrical  Magnitudes    are    lines,    surfaces,  arid 
solids. 

3.     Geometrical  Magnitudes. 


D.  Solid. 

c. 

Surface. 

w 

A.  Lin 

&. 

Oi  4^ 

ri 
P 

03 

0 

0 
3 
ft 

2.  Cylinder. 

,1 

3 

0 

i 

,  —  *— 

4*.  00  10 

<t 

r-»        4^  CO  t-0  HJ 

3 

9 

^d  c/5   o 

tTp 

003 
C    ? 

<  S 

—^a'a- 
J~"    ,^- 

/^A^S 

to  i-1 

to  i- 

-i  Oi  4^  03  to 

^     SPOO 

co     RpwO^d 

B§:«| 

5 

OK 

0? 

0  £  0  0* 

P  ^  o  ' 

"g.    P  S^"^ 

^> 

N 

I  p-'cS 


?§asj?p??^a 

'     »   p    — ^  '     X   3   »  P 

as^ciS      »S"CL3 

s«ss-  |«s<§ 
tUB 


H->  00  bO  H-1  tO  H* 


3  t 
arq  P 

c  r« 


o 
p 

I 

o 

I 


cr—  » 


5d  ^ 

3"  g- 

I  1 

i— *  CTQ 

I  " 

51  = 


MENSURATION.  193 

A  Line  is  a  geometrical  magnitude  having  length,  with- 
out bread! h  or  thickness. 

5.  A  Straight  Line  is  a  line  which  pierces  space  evenly, 
so  that  a  piece  of  space  from  along  one  side  of  it  will  fit  any  side 
of  any  other  portion. 

6.  A  Curved  Line  is  a  line  no  part  of  which   is  straight. 

7.  A    Surface  is  the  common   boundary  of  two  parts  of  a 
solid,  or  of  a  solid  and  the  remainder  of  space. 

8.  A  Plane  Surface,  or  Plane,  is  a   surface   which  di- 
vides space  evenly,  so  that  a  piece  of  space  from  along   one  side 
of  it  will  fit  either  side  of  any  other  portion  of  it. 

9.  A  Ctirved  Surface  is  a   surface    no    part   of  which  is 
plane. 

10.  A    Polygon       (ttMyatvos,     from     Ilokuq^     many,     and 
angle)    is   a   portion   of  a   plane   bounded   by   straight  lines. 


11.  A  Circle  (xtpxos,  circle,  ring)  is  a  portion  of  a  plane 
bounded  by  a  curved  line  every  point  of  which  is  equally  distant 
from  a  point  within  called  the  center, 


1£.  An  Ellipse  (e'>Ue£0££)  is  a  portion  of  a  plane  bounded 
by  a  curved  line  any  point  from  which,  if  two  straight  lines  are 
drawn  to  two  points  within,  called  the^/bcz,  the  sum  of  the  two 
lines  will  be  constant. 

13.  A    Triangle    (Lat.      Triangulum,     from    tries,    tria, 
three,  and  angulus,  corner,  angle)  is  a  polygon  bounded  by  three 
straight  lines. 

14.  An  Angle  is   the   opening   between  two   lines   which 
meet  in  a  point. 

II.   Straight  Angle. 
8.  Oblique   j  I  Acute^ 

16.  A  Straight  Angle  has  its  sides  in  the  same  line,  and 
on  different  sides  of  the  point  of  meeting,  or  vertex. 

17.  A  Might  Angle  is   half  of  a    Strait  Angle,   and   is 
formed   by  one   straight  line  meeting  another  so  as  to  make  the 
adjacent  angles  equal. 

18.  An   Oblique  Angle  is  formed  by  one   line  meeting 
another  so  as  to  make  the  adjacent  angles  unequal. 

19.  An,  Acute  Angle  is  an  angle  less  than  a  right  angle. 

20.  An   Obtuse  Angle  is  an   angle   greater  than  a  right 
angle. 


194  FINKEL'S  SOLUTION  BOOK. 

21*  A  Hight  Triangle  is  a  triangle,  one  of  whose  angles 
is  a  right  angle. 

22.  An  Oblique-  Angled  Triangle  is  one  whose  angles 
are  all  oblique. 

23.  An  Isosceles  Triangle  is  one  which  has  two  of  its 

sides  equal. 

24.  A  Scalene  Triangle  is  one  which  has  no  two  of  its 

sides  equal. 

25.  An   Equilateral  Triangle  is  one  which  has  all 
the  sides  equal. 

26.  A   Quadrilateral  (Lat.  quadrilaterus  ,  from  quatuor^ 
four,  and  latus,  lateris,  a   side)  is  a  polygon  bounded   by  four 
straight  lines. 

21.  A  Parallelogram  (  flap  ally  16?  pawov,  from 
IlapdMyJios,  parallel,  and  rpafi.fjt.rj,  a  stroke  in  writing,  a  line) 
is  a  quadrilateral  having  its  opposite  sides  parallel,  two  and  two. 

28.  A  Hight  Parallelogram  is  a  parallelogram  whose 
angles  are  all  right  angles. 

29.  An   Oblique  Parallelogram  is  a  parallelogram 
whose  angles  are  oblique. 

30.  A    Rectangle   (Lat.   rectus,   right,   and    angulus,   an 
angle)  is  a  right  parallelogram. 

31.  A   Square  is  an  equilateral  rectangle. 


32.     A  ^Rhomboid     (fafjLfioetd&s,  from  ^o^/foc,  rhomb,   and! 
shape)  is  a  parallelogram  whose  angles  are  oblique. 


33.  A  Rhombus     fitppos,   from     jMpfieiv,  to  turn  or  whirl 
round)  is  an  equilateral  rhomboid. 

34.  A  Pentagon        (  Ilsvrd^wvov,     /7^re,      five,      and     futvia> 
angle)  is  a  polygon  bounded  by  five  sides.     Polygons  are  named 
in  reference  to  the  number  of  sides  that  bound  them.     A  Hexa- 
gon has  six  sides;  Heptagon,  seven;   Octagon,  eight;  Nonagonr 
nine;    Decagon,    ten;    Undecagon,    eleven;    Dodecagon,    twelve; 
Tridecagon,   thirteen  ;     Tetradecagon,   fourteen;    Pentedecagon, 
fifteen;  Hexdecagon,  sixteen;    Heptadccagon,    seventeen;     Octa- 
decagon,  'eighteen;    Enneadecagon  ,   nineteen;  Icosagon,  twenty; 
Icosaisagon,  twenty-one  ;  Icosad&agon,  twenty  -two;  Icosatriagon, 
twenty-three  \Icosatetragon,  twenty-four  ;  Icosapentegon  ,  twenty- 
five;     Icosakexagon,     twenty-six;    Icosaheptagon,   twenty-seven; 
Icosaoctagon,  twenty-eight  ;    Icosaenneagon,  twenty-nine;    Tria- 
contagon,  thirty;     Tficontaisagon,  thirty-one;     Tricontadoagon, 
thirty-two;    Tricontatriagon,  thirty-three;  and    so    on   to  Tessa- 
racontagon,   forty;    Pentecontagon,    fifty;    Hexacontagon,   sixty;: 


MENSURATION.  195 


Hebdomacontagon ,  seventy  ;  Ogdoacontagon,  eighty ;  Enenacon- 
tagon,  ninety;  Hecatonagon ,  one  hundred;  Diacosiagon,  two 
hundred;  Triacosiagon,  three  hundred;  Tetracosiagon,  lour  hun- 
dred; Pentecosiagon,  five  hundred;  Hexacosiagon,  six  hundred; 
Heptacosiagon,  seven  hundred  ;  Oktacosiagon,  eight  hundred  ; 
Enacosiagon,  nine  hundred;  Chiliagon,  one  thousand;  &c. 

35.  A  Spherical  Surface  is   the  boundary  between  a 

sphere  and  outer  space. 

36.  A    Conical   Surface  is  the  boundary  between  a  cone 
and  outer  space. 

37.  A  Cylindrical    Surface   is   the   boundary  between 
the  cylinder  and  outer  space. 

38.  A  Solid  is  a  part  of  space  occupied  by  a  physical  body, 
or  marked  out  in  any  other  way. 

39.  A  Polyhedron    (IJoMedpoq,     from    77oAy^,   many,  and 
seat,  base)  is  a  solid  bounded  by  polygons. 

40.  A  Prism  is  a  polyhedron  in   which   two   of  the  faces 
are  polygons  equal  in  all  respects  and   having   their  homologous 
sides  parallel. 

41.  The  Altitude  of  a  prism  is  the  perpendicular  distance 
between  the  planes  of  its  bases. 

4:2.     A  Triangular  Prism  is  one  whose  bases  are  trian- 
gles. 

43.  A  Quadrangular  Prism  is  one  whose  bases  are 
quadrilaterals. 

44.  A  Parallelopipedon   is    a    prism    whose*  bases  are 
parallelograms. 

45.  A  Hight   Parallelopipedon  is  one  whose   lateral 
edges  are  perpendicular  to  the  planes  of  the  bases. 

46.  A  Rectangular  Parallelopipedon  is  one  whose 
faces  are  all  rectangles. 

47.  A  Cube    x6/5o<?,     a  cube,  a  cubical  die)  is  a  rectangular 
parallelopipedon  whose  faces  are  squares. 

48.  A  Hight  Prism  is  one  whose  lateral   edges   are  per- 
pendicular to  the  planes  of  the  bases. 

49.  An  Oblique  Prism  is   one   whose   lateral   edges  are 
oblique  to  the  planes  of  the  bases. 

50.  A  Pyramid  (flupa^)  is  a  polyhedron  bounded  by  a 
polygon  called  the  base,  and  by  triangles  meeting  at  a  common 
point  called  the  vertex  of  the  pyramid. 


196  FINKEL'S   SOLUTION  BOOK. 

51.     The  Convex  Surface  of  a  pyramid  is  the  sum  of  the 

triangles  which  bound  it. 

58.  A  Hight  Pyramid  is  one  whose  base  is  a  regular 
polygon,  and  hi  which  the  perpendicular,  drawn  from  the  vertex 
to  the  plane  of  the  base,  passes  through  the  center  of  the  base. 
The  perpendicular  is  called  the  axis. 

53.  A  Tetrahedron     (rirpa    four,    and    Ifl/oa,     seat,  base) 
is  a  pyramid  whose  faces  are  all  equilateral  triangles. 

54.  The  Altitude  ( Lat.     Altitudo,  from  altus,   high,  and 
ude  denoting  state  or  condition)  of  a  pyramid  is  the  perpendicu- 
lar distance  from  the  vertex  to  the  plane  of  the  base. 

55.  The  Slant  If  eight  of  a  pyramid,   is   the  perpendicu- 
lar distance  from  the  vertex  to  any  side  of  the  base. 

56.  A  Triangular  Pyramid  is  one  whose  base  is  a 
triangle. 

57.  An    Octahedron       (dxrdsdpos     from    dxrtf     eight,  and 
tSpa     seat,  base)  is  a  polyhedron  bounded  by   eight  equal  equi- 
lateral triangles. 

58.  A   Dodecahedron     (dudsza,      twelve,  and  gdpa,   seat, 
base)    is  a  polyhedron    bounded   by    twelve    equal    and    regular 
pentagons. 

59.  An  Icosahedron  (efxo?*,  twenty,  and  Zdpa,  seat,  base) 
is  a  polyhedron  bounded  by  twenty  equal  equilateral  triangles. 

60.  A  Cylinder       (xuhvdpoz,    from    xuMvdew,    xvUew,    to 
roll)  is  a  solid  bounded  by  a  surface  generated  by  a  line  so  mov- 
ing that  every  two  of  its  positions  are  parallel,  and   two  parallel 
planes. 

61.  The  Axis    (a^utv)    of  a  cylinder  is  the  line  joining  the 
centers  of  its  bases. 

62.  A  Hight  Cylinder  is  one  whose  axis   is  perpendicu- 
lar to  the  planes  of  the  bases. 

63.  A    Cone    (xwvos,    from    Skr.  co,  to  bring  to  a   point)  is 
a  solid  bounded  by  a  surface  generated  by  a  straight  line  moving 
so  as  always  to  pass  through  a  fixed  point  called  the  apex,,  and  a 
plane. 

64.  A  Hi(/ht  Cone   is   a   solid   generated  by   revolving  a 
right-angled  triangle  about  one  perpendicular. 

65.  An  Oblique  Cone  is   one   in    which   the   line,  called 
the  axis,  drawn  from  the  apex  to    the   center   of  the   base   is  not 
perpendicular. 

66.  The  Frustum  (Lat.  frustum,  piece,  bit)    of  a  pyra- 
mid or  a  cone  is  the    portion   included    between   the   base   and  a 
parallel  section. 

67.  A  Sphere    (ff^aipa)    is  a   solid  bounded  by  a  curved 


MENSURATION. 


197 


surface,  every    point   of  which  is   equally   distant   from   a  point 
within,  called  the  center. 

Before  we  enter  into  the  solution  of  problems  in  Mensuration, 
it  will  be  necessary  first  to  explain  a  difficulty  which  we  en- 
counter. 

The  common  way  of  teaching  that  feet  multiplied  by  feet  give 
square  feet  is  wrong  ;  for  there  is  no  rule  in  mathematics  justify- 
ing the  multiplication  of  one  denominate  number  by  another. 
If  it  is  correct  to  say  feet  multiplied  by  feet  give  square  feet,  we 
might,  with  equal  propriety,  say  dollars  multiplied  by  dollars 
give  square  dollars — a  product  wholly  unintelligible.  In  all  our 
reasoning,  we  deal  with  abstract  numbers  alone  or  the  symbols 
of  abstract  numbers.  These  do  not  represent  lines,  surfaces,  or 
solids,  but  the  relations  between  these  numbers  may  represent 
the  relations  between  the  magnitudes  under  consideration. 

Suppose,  for  example,  that   the  line  AB  contains  5  units,  and 
the  the  line  BC  4  units.     Let  a  denote   the   abstract  number  5, 
and  b  the    abstract  number  4.     Then 
#3=20.     Now  this  product  ab   is  not  **' 
a   surface,   nor  the   representation    of 
a   surface.     It  is   simply  the    abstract 
number    20.     But  this   number   is  ex- 
actly   the    same   as     the     number    of 
square  units  contained  in  the  rectan- 
gle whose  sides  are  AB  and  B  C,  as 
may  be  seen  by  constructing  the  rec-  A 
tangle  AB  CD.     Hence  the  surface  of 
the  rectangle  is  measured  by  20  squares  described  on  the  unit  of 
length. 

This  relation  is  universal,  and  we  may  always  pass  from  the 
abstract  thus  obtained  by  the  product  of  any  two  letters,  to  the 
measure  of  the  corresponding  rectangle  by  simply  considering 
the  abstract  units  as  so  many  concrete  or  denominate  units. 

In  like  manner,  the  product  of  three  letters  abc  is  not  a  solid 
obtained  by  multiplying  lines  together,  which  is  an  impossible 
operation.  It  is  simply  the  product  of  three  abstract  numbers 
represented  by  the  letters  «,  b,  and  c,  and  is  consequently  an 
abstract  number.  But  this  number  contains  precisely  as  many 
units  as  there  are  solid  units  in  the  parallelopipedon  whose  edges 
correspond  to  the  lines  a,  b,  and  c\  hence,  we  may  easily  pass 
from  ths  abstract  to  the  concrete.  Hence,  if  we  wish  to  find  the 
area  of  a  rectangle  whose  width  is  4  feet  and  length  6  feet,  we 
simply  say,  6x4=24  square  feet.  We  pass  at  once  from  the  ab- 
stract in  the  first  member  to  the  concrete  in  the  second. 

It  is  a  question  whether  pupils  should  be  taught  a  falsehood 
in  order  that  they  may  learn  a  truth. 

(See  Bledsoc's  Philosophy  of  Mathematics,  pp.  97-106.) 


198  FINKEL'S  SOLUTION  BOOK. 


I.     PARALLELOGRAMS. 

Prob.  I.    To  find  the  area  of  a  parallelogram  ;  whether  it 
be  a  square,  a  rectangle,  a  rhomboid,  or  a  rhombus. 


.  —  ^4  =  /X^,  where  ^4=area,  /—length,  and  &= 
breadth  ;  or,  A—bX&,  where  yl=area,  £=base,  and  tf=altitude. 

Rule  —  Multiply  the  length  by  the  breadth;  or,  the  base  by  the 
altitude. 

I.     What  is  the  area  of  a   parallelogram    whose   length    is  15 
feet  and  breadth  7  feet? 

By   formula,   ^4=/x£=lengthXbreadth=15X  7=105   sq. 
feet. 

[  1.  15  feet=length. 
TT  1  2.  7  feet=breadth. 
H8.  .'.  15X7=105  sq.  ft. 
=area. 

III.     .-.  The  area    i  s  105   sq. 
ft.  .......   _ 

""  FIG.  4. 

Note.  —  The  base  is  not  necessarily  the  side  toward  the  ground.  Thus  in 
the  parallelogram  ABCD,BC  may  be  considered  the  base,  in  which  case, 
the  altitude  would  be  the  perpendicular  distance  EF,  between  the  sides  BC 
and  AD.  If  HG  and  B  C  were  given,  we  could  not  find  the  area  of  the  par- 
allelogram because  we  have  not  the  base  and  altitude  given. 

I.     What  is  the  area  of  the  parallelogram  ABCD,   if  BC  is 
26  feet  and  EFW  feet? 


By  formula,  ^=ax£=^^X/?C=50X  26=1300  sq.  ft. 

II.  26  feet=,#C=  base. 
2.  50  feet=-E,F=altitude. 
3.  /.  26X50=1300  sq.  ft.=area.      . 
III.     .-.  The  area  of  ABCD=V&W  sq.  ft. 

I.     A  floor  containing  132  square  feet,  is  11  feet  wide  ;    what 
is  its  length? 


By  formula,  A^l^b.     .-.  /=^4-f-£=132-f-ll=12  ft. 

II.   132  sq.  ft.=area. 
2.  11  ft.=breadth. 
3.  132-:-!  1=12  ft.=length. 

III.     /.  The  floor  is  12  ft.  long. 


Prob.  II.    The  diagonal  of  a  square  being  given,  to  find 
the  area. 

Formula. — A—d  2-i-2. 

Rule. — Divide    the   square  of    the    diagonal   by   2,    and  the 
quotient  will  be  the  area. 

I.     What  is   the  area  of  a  square 
whose  diagonal  is  8   chains? 

By  formula,  ^4=<f  2-7-2  =  82-r- 
2=32  sq.  chains. 

1.  8  ch.=length  of  diagonal= 
BD. 

II.)          square  described  on  the  di- 
agonal BD. 

3.  32  sq.  ch.=64   sq.    ch.-f-2= 
area  of  the  square  AB  CD. 

III.  V.%  32  sq.  ch.=the  area  of  the  FIG.  5. 

square. 

Prob.  III.    The  area  of  a  square  being  given,  to  find  its 
diagonal. 

Formula. — d=\/^A. 

. — Extract  the  square  root  of  double  the  area. 
The  area  of  a  square  is  578  sq.  ft.  ;  what  is   the  diagonal? 
By  formula,  d=\/ '2A=\/2Xo7 '8= V 'l!56=34  feet. 

1.  578  sq.  ft.=area  of  the  square. 

2.  1156  sq.  ft.=2x578  sq.  ft.=double  the  area. 

3.  34  feet=\/Ti56=the  diagonal. 
.-.  The  diagonal  is  34  feet. 


II. 


III. 


Prob.  IV.    The  diagonal  of  a  square  being  given,  to  find 
its  side. 


Formula.  —  -S^ 

Rule.  —  Extract  the  square   root  of  one-half  the  square  of  the 
diagonal. 

I.     What  is  the  side  of  a  square  whose  diagonal  is  12  feet? 
By    formula,    S=y'p2=v'l>^=V^=6\/2=8.4852+ft 

II.   12  ft.=the  diagonal. 
2.  144  sq.  ft.=122=square  described  on  the  diagonal. 
3.  72  sq.  ft.—  area  of  square  whose  side  is  required. 
4.  .-.  8.4852  ft.==6\/2==\/72=side  of  the  square. 
III.     /.  The  side  of  the  square  is  8.4852+ft. 


200  FiNKEL'S   SOLUTION   BOOK. 

Prob.  V.    To   find   the   side   of  a   square   having-  its  area 
given. 


Rule.  —  Extract  the  square   root  of  the  number  denoting  its 
'area. 

I.     What  is   the   side   of  a  square   field   whose   area   is   2500 
square  rods? 

By  formula,  S=v^=V250()=50  rods. 

(  1.  2500  sq.  rd.=area  of  the  field. 

|  2.  50  rd.=Vr2500=side  of  the  square  field. 

III.     .;.  The  side  of  the  field  is  50  rods. 

II.  TRIANGLES. 

Prob.  VI.    Given  the  base  and  altitude  of  a  right-angled 
triangle,  to  find  the  hypothenuse. 


Formula.  —  /£= 


5. —  To  the  square  of  the  base  add  the  square  of  the   alti- 
tude and  extract  the  square  root  of  the  sum. 

I.     In  the  right  angled  triangle  A  CB ',  the  base   A  C=56  and 
the  altitude  /?C=33  ;  what  is  the  hypothenuse? 

By    formula,   ^=\/rfz+*58=V33*+56*=\/1089+31bt)=Vi225 
=65. 

1.  56=^4  C=the  base. 

2.  3136=562=the    square   of  the 

base. 

3.  33=^C=the  altitude. 

4.  1089=338=the    square    of    the 

altitude. 

5.  4225=3 1364rl089=the  sum  of 

the  squares  of  the  base  and  altitude. 

6.  65— \/ 4225=  the  square  root  of  the  sum  of  the  squares  of 
I          the  base  and  altitude=the  hypothenuse. 

III.      .-.  The  hypothenuse=65. 

Prob.  VII.    To  find  a  side,  when  the  hypothenuse  and  the 
other  side  are  given. 


II. 


Formula*.-      = 

f  h— 


Rule.  —  From  the  square  of  the  hypothenuse  subtract  the  square 
of  the  given  side  and  extract  the  square  root  of  the  remainder. 


MENSURATION. 


201^ 


I.      The    hypothenuse  of    a  right-angled    triangle  is   109,  and 
the  altitude  60;  what  is  the  base? 


By  formula,  b—^/~k'1—  ^=\/1092—  002=A/8281=91. 

1.  109=hypothcn.use. 

2.  11881=1092=square  of  the  hypothenuse. 

3.  S0=the  altitude. 

II  <  4.  3600=602=the  square  of  the  altitude. 

5.  8281=11881—  3600=difference    of   the    squares    of   the 

hypothenuse  and  altitude. 

6.  9  1=  'V/8281=the  square  root  of  this  difTerence=the  base. 

III.     .'.  The  base  is  91. 

Remark.  —  When  a=6,  >h=rV/2a2=rt\//2.     From  this,  we  see  that  the  diag- 
onal of  a  square  is  v  2  times  its  side. 

Prob.  VIII.    To  find  the  area  of  a  triangle,  having-  given 
the  base  and  the  altitude. 

Formula.  —  A=\a  x  b. 

Rule.  —  Multiply  the  base  by   the  altitude  and  take  half  the 

product. 

I.      What  is  the  area  of  a  triangle  whose    base    is   24    feet  and 
altitude  16  feet? 

By  formula,  A=\a^b=\  X  16x24=192  sq.  ft. 

1.  24  ft.=base. 

2.  16  ft.=altitude. 

,3.  384  sq.  ft.=16  X24 
IIJ  duct  of  base  and    altitude. 

4.  192  sq.  ft.=i  of  3S4  sq.  ft.== 

half  the  product  of  the  base 
and  the  aititude=area. 

III.     .•.  The  area  of  the   triangle  is 
192  sq.  ft. 


FIG.  7. 


Prob.  IX.    To  find  the  area  of  a  triangle,  having  given 
its  three  sides. 


Formula.  —  A  = 


—a  )  (  .v_^  )  (  x—c  )  ,  where  s=^  (a-\-b-\-c). 

—  Add  the  three  sides  together  and  take  half  the  sum; 
from  the.  half  sum,  subtract  each  side  seperately;  imdtiply  the  half 
sum  atid  the  three  remainders  together  and  extract  the  square  root 
of  the  product. 

*  Demonstration.  —  ^In  Fi<?.  7,  let  A  C=.b,  BC-=.a,  and  AB=^c.  In  the 
right-angled  triangle  ADB.  I3DZ—AB''~~  AD*,  and  in  the  nefht-angled 
triangle  CDB,  BD*=BC*—DC*.  .'.  AB*—AD'*=BC'i—DC'i.  or  r2— 


202  FINKEL'S   SOLUTION  BOOK 

I.     What  is  the  area  of  a  triangle  whose  sides  are  13,   14,  and 
15,  feet  respectively  ? 

By   formula,  A=\/s(s—a)(s—b)(s—7)=*j21X  (21—13)  X  (21— 
14)X(21— 15)  =V/21X  8X7X6  =  \/7056=84  sq.  ft. 

'l.  42  ft.=13  ft.+14  ft.+15  ft.=sum  of  the  three  sides. 

2.  21  ft.=4  of  42  ft.=half  the  sum  of  the  three  sides. 

3.  21  ft. — 13  ft.=8  ft.=first  remainder. 

n  ,4.  21  ft— 14  ft.=7  ft.=second  remainder. 

^5.  21  ft. — 15  ft.=6  ft.=third  remainder.  [mainders, 

6.  7056=21  X6X 7  X8=product  of  half  sum   and   three  re» 

7.  84  sq.  ft.=  \/7056=square  root  of  the  product  of  the  half 

sum  and  three  remainders=the  area  of  the  triangle. 

III.     .-.  The  area  of  the  triangle  is  84  sq.  ft. 

Prob.  X.    To  find   the   radius   of  a   circle   inscribed  in  a 
triangle. 

Formula.-— R=&A-r-(a+b-\-c). 

*Rule. — Divide  twice  the  area  of  the   triangle  by  the  sum  of 
the  three  sides. 

I.     Find  the  radius  of  a  circle  inscribed    in   a    triangle  whose 
;sides  are  3,  4,  and  5  feet,  respectively. 

1.   6  sq.  ft.=  V-y(j — a)(s — b)(s — c)=area   of  the    triangle, 
by  formula,  Prob.  IX. 


II. 


2.   12  sq.  ft.— twice  the  area  of  the  triangle. 


3.  12  ft.=3  ft+4  ft.+  5  ft.=sum  of  the  three  sides. 

4.  .*.  1  ft.=12-f-12=twice  the  area  divided  by   the   sum  of 

the  sides—  the  radius  of  the  inscribed  circle. 
III.      .-.  The  radius  of  the  inscribed  circle  is  1  ft. 

AD2=a2—  DC2,  whence  c2—  a2—  AD2—  DC2.  But  AD2—DC2=(AD 
+DC^(AD—DC)=b(AD—DC).  .'.  b(AD—DC)=c2—a2,and  AD—  DC 
=(c2  —  «2)-f-£.  But  AD-\-DC=b.  .'.  By  adding  the  last  two  equations,  we 


have     2AD=  -  =  --  h^=  -  T  -  ;     whence    A£>=  -  —  -  -  '        Since 
oo  .  2o 


!,  if  we  substitute  the  value  of  AD   just  found, 
(r*—a*+-- 
we  have 


=c!-(--^)2--    ^ 

-b2 )  (2bc— c2+a2— b2 )_  (b2  -f  2bc+c2- 


(2Z>c+c2—a2+l>2  )  (2bc—  c2-}-  a2—  b2  )_  (b2  +  2t>c+c2—a2  )[a^—(^—2bc-}-c2  )  ] 

4b2  4P 

[(b+c)2—  a2][a2—  (b—  c}2} 


Now   the 
area  of  ABC—\A  CKBD. 

-—    2] 


%y—a)(j— -£)(j— e),  where  2j=(a-f*-f-c).     Q.  E.  D. 
*Note. — For  Demonstration,  see  any  geometry. 


MENSURATION.  203 

Prob.  XI.     To  find  the  radius  of  a  circle,  circumscribed 
about  a  triangle  whose  sides  are  given. 

abc  abc 

Formula. — /?= — -=- — .  f 

4A     4  \s(s— a)(s— b)(s— c). 

*Rule. — Divide  the  product  of  the  three  sides  by  four  times  the 
•area  of  the  triangle. 

I.     What  is  the  radius  of  a   circle   circumscribed   about   a  tri- 
angle whose  sides  are  13,  14,  and  15  feet,  respectively? 

1.  2730  cu.  ft.=13Xl4Xl5=the  product  of  the  three  sides. 

2.  84  sq.   ft.=  *ls(s— a)(s— b)(s — c)=the   area   of  the  tri- 

angle, by  Prob.  IX.  [angle. 

II.<[3.  336  sq.  ft.=4x84  sq.  ft.=four  times  the  area  of  the  tri- 
4.  8J-  ft.=2730-7-336=the  product  of  the  three  sides  divid- 
ed by  four  times  the  area  of  the  triangle=the  radius 
of  the  circumscribed  circle 

III,      .•.  The  radius  of  the  circumscribed  circle  is  8-|  ft. 

Prob.  XII.      To  find  the  area  of  an  equilateral  triangle, 
having  given  the  side. 

Fomiwlct- — -A=-z  V3.S"2 ,  where  s=side.     This  is  what  Prob. 
IX.  becomes,  when  a=b=c. 

Rule.— Multiply  the  square  of  a  side  by  %  V3,=-433013+. 

I.     What  is  the  area  of  an   equilateral   triangle   whose   sides 
are  20  feet? 

* Demonstration. — Let  ABC  be  any  triangle,  and  ABCE  the  circumscrib- 
ed circle.  Draw  the  diameter  BE,  and 
draw  EC.  Draw  the  altitude  BD  of  the 
triangle  ABC.  The  triangles  ADB  and 
BCE  are  similar,  because  both  are  right- 
angled  triangles,  and  the  angle  £*AD=the 
angle  BEC.  Hence,  AB'-EB'-  '-BD'-BC. 
Hence,  AB^BC—BE^BD  or  ac=2RX 
BD.  But,  in  the  demonstration  of  Prob. 

2 
IX.,   we    found    BD=-j-\/S(8—a)(8—b)(s—c). 

Whence    

UK 

18. 


204  FINKEL'S   SOLUTION  BOOK. 

By  formula,  A=%  V3X202=100  V3=173.205+sq.  ft. 

II.  20  ft.=length  of  a  side. 
2.  400  sq.  ft.=202=square  of  a  side. 
3.  173.205   sq.    ft.=t  V3X400=.433013X400=i\/3  times 
the  square  of  a  side,=the  area  of  the  triangle. 

III.     .'.  The  area  of  the  equilateral  triangle  is  173.2054-sq.  ft. 

Prob.  XIII.  The  area  and  base  of  a  triangle  being  given, 
to  cut  oil  a  triangle  containing-  a  given  area,  by  a  line  run- 
ning parallel  to  one  of  its  sides. 

12' 

Formula. —  b'—b^—r-,  where    A=area    of    the     given 

triangle;  b^  the  base  of  the  given  triangle;  and  A  ,  the 
area  of  the  portion  to  be  cut  off. 

Rule. — As  the  area  of  the  given  triangle  is  to  tne  area  of  the 
triangle  to  be  cut  off,  so  is  the  square  of  the  given  base  to  the 
square  of  the  required  base.  The  square  root  of  the  result  "Mill 
be  the  base  of  the  required  triangle. 

I.  The  area  of  the  triangle  ABC  is  250  square  chains  and 
the  base  AB,  20  chains  ;  what  is  the  base  ,of  the  trian- 
gle, area  equal  to  60  sq.  chains,  cut  off  by  ED  parallel 
to 


II. 


By  formula,  ^4Z>==3>=20N=4V6  =-9.7979  +  ch 

'1.  250  sq.  ch.=area  of  the  given  triangle  ABC. 

2.  60    sq.    ch.=area    of   the 

triangle  A  JED. 

3.  20  ch.=base  of  the  trian- 

gle ABC. 

4.  .-.  250  sq.  ch.  :  60  sq.  ch. 

::  203  :  AD*.     Whence 

5.  A£>2=  (400X60)  -^-250 

=-96. 
,6.  .;.  ^Z^V96=9.7979+ch.  FIG,  9. 

III.     .-.  The  base  ^4Z>=9.7979+ch. 

III.  TRAPEZOID. 

Prob.  XIV.     To  find  the  area  of  a  trapezoid,  having-  given 
the  parallel  sides  and  the  altitude. 


*  —  A—^(b-\-b')a,  where  b  and  b'  are  the  parallel 
sides  and  «,  the  altitude. 

Rule.  —  Multiply  half  the  sum  of  the  parallel  sides  by  the  al- 
titude. 


MENSURATION. 


205 


II. 


I.  What  is  the  area  of  a  trapezoid  whose  parallel  sides  are 
15  meters  and  7  meters  and  altitude  6  meters? 

By  formula,  A=\(b+b')  x«=^(15+7)  X6=66  m8. 
•I.  7  m,=Z>C,the  length  of  one 
of  the  parallel  sides,  and 

2.  15  m.=AB,  the  length  of 

the  other  side. 

3.  22  m.=7  m.-(-15   m.=sum 

of  the  parallel  sides. 

4.  11    m.=4  of  22    m.=half 

the  sum   of  the   parallel 
sides.  FIG.  10. 

.5.  66  m2.=6xll=area  of  the  trapezoid,  ABCD. 

III.      .'.  The  area  of  the  trapezoid  is  66  m2. 

IV.  TRAPEZIUM  AND  IRREGULAR  POLYGONS. 

Prob.  XV.     To  find  the  area  of  a  trapezium  or  any  irreg- 
ular polygon. 

Rule, — Divide  the  figure  into  triangles ,  find  the   area   of  the 
triangles  and  take  their  sum. 

I.  What  is  the  area  of  the  trapezium  ABCD,  whose  diag- 
onal A  C  is  84  feet,  and  the  perpendiculars  D  JB  and 
BF,  56  and  22  feet,  respectively? 

1.  84  ft.=^C=Aase  of    the 

triangle  ADC. 

2.  56    ft.=D£  =  altitude  of 

ADC. 

3.  /.   2352   sq.    ft.=i(^4CX 

D-£')=area  of  the  trian- 
gle AD  C. 

4.  84   ft=^4C=base   of  the 

triangle  ABC. 

5.  22  &.=£!?==  altitude   of 

ABC. 

6.  .•.924sq.ft.=^(^4CX^^')  FIG.  11. 

=area  of  the  triangle  ABC. 

7.  3276  sq.  ft.=2352  sq.  ft. +924  sq.   ft.=ADC+A£C= 

area  of  the  trapezium  AB  CD. 

III.     .-.  The  area  of  the  trapezium  ABCD  is  3276  sq.  ft. 

V.  REGULAR  POLYGONS. 
Prob.  XVI.    To  find  the  area  of  a  regular  polygon. 

. — A=\aY,pi  where  p  is  the  perimeter  and  #, 
the  apothem. 

5. — Multiply  the  perimeter  by  half  the  apothem. 


206 


FINKEL'S  SOLUTION  BOOK. 


The  Perimeter  of  any  polygon  is  the  sum  of  all  its  sides. 

The  Apothem  is  the  perpendicular  drawn   from   the  cente*  to 

any  side  of  the  polygon. 

I.     What  is  the  area   of  a  regular  heptagon   whose  slJe  is 
19.38  and  apothem  20? 

1.  19.38=length  of  one  side. 

2.  135  66=length  of  7  sides=the  perimeter. 

3.  20=apothem. 

4.  10=^  of  20=half  the  apothem. 

5.  1356.6=10Xl35.66=product  of  perimeter  b>   half  the 

apothem. 

III.     .-.  The  area  of  the  heptagon  is  1356.6. 

Prob.  XVII.  To  find  the  area  of  a  regular  poiygon.  when 
the  side  only  is  given. 

*Rllle. — Multiply  the  square  of  the  side  of  the  -polygon  by  the 
number  standing  opposite  to  its  name  in  the  following  table  of 
areas  of  regular  polygons  whose  side  ^*  1  : 


Name. 

Sides. 

Multipliers. 

Triangle, 

3 

£V§              =     .4330127. 

Tetragon,  or  square, 

4 

1                 =  1.0000000. 

Pentagon, 

5 

fVl-l  |V5    =  1.7204774. 

Hexagon, 

6 

4V3             =  2.5980762. 

Heptagon, 

7 

4  cot.  if00^  3.6339124. 

Octagon, 

8 

2-f  2V§        =  4.8284271. 

Nonagon, 

9 

f  cot.  20°  =  6.1818242. 

Decagon, 

10 

IV/5+2V5    =  7.6942088. 

Undecagon, 

11 

Vcot.  W°°=  9.3656399, 

Dodecagon, 

12 

3(2+Vs)     =11.1961524, 

*  Demonstration. — Since  a  regular  polygon  can  be  divided  into  as  many 
equal  isosceles  triangles  as  it  has  sides,  we  may  find  the  area  of  one  trian- 
gle and  multiply  this  area  by  the  number  of  triangles,  for  the  whole  area. 
Let  yl^C  be  one  of  these  isosceles  triangles,  taken  from  a  polygon  of  n 
s;des,  AB  and  BC  the  equal  sides,  and  A  C  the  base.  The  angle  at  the  ver- 
tex 5=360°-i-».  ^4=K180°— 360°-5-«)—  C.  From  B  let  fall  a  perpendicu- 

/?  7~)  1 80° 

lar  on  A  C  at  D.     Then  by  trigonometry,  —— =tan  (90° ).      .'.  BD=z 

\-A.  C 


(1 80°  \ 
— —  1.     The  area  of  the  triangle  ABC~\ACKBD=.\AC*  cot 

(_ V     .'.  The  area  of  the  polygons:— A  C2  cot  (  — —  )r=-.y*  cot   ( | 
n    )                                                          4                  \^/4               \»/ 

where ^=side.     By  placing  s=l,  and  ^=13,  14,  15,  &c.,  respectively, the  area 
of  polygons  of  13,  14,  15,  &c.,  side  respectively,  may  be  found. 


MENSURATION.  207 


Prob.  XVIII.    To  find  the  side  of  an  inscribed  square  of  a 
triangle,  having"  given  the  base  and  the  altitude. 

Formula. — 9= — r- /,   where  s=side,  b  the  base,   and  a 
a-\-b 

the  altitude. 


'• 

*Rule 


*  —  Divide  the  product  of  the  base  and  altitude  by  their 
sum, 

I.     What  is  the  side  of  an   inscribed   square  of  a   triangle 
whose  base  is  14  feet  and  altitude  8  feet? 

.  ab       14X8 

By  formula,  s=—  =^—=5^  feet. 

1.  8  feet=the  altitude. 

2.  14  feet=the  base. 

II.<j3.  112  sq.  ft.=14x8=the  product  of  the  base  and  altitude. 
4.  22  feet=14  ft.-f  8  ft.=their  sum. 
.5.  SyV  feet=112-7-22=the  product  divided  by  the  sum 


III.     .'.  5Ty  ft.=the  side  of  the  inscribed  square. 

VI.  CIRCLE. 

Rrob.  XIX.      To  find  the  diameter  of    a  circle,  having- 
given  the  height  of  an  arc  and  a  chord  of  half  the  arc. 

Formula.  —  D=k2-±-a,  in  which  £=chord   of  half  the 
arc  and  0:±=height. 

t  Rule.  —  Divide  the  square  of  the  chord  of  half  the  arc  by  the 
height  of  the  chord. 

*  Demonstration.  —  Let  ABC  be  any  triangle  whose  base  is  b  and  altitude 
a.  Produce  AC  to  H,  making  CH—BD.  At  H,  erect  the  perpendicular 
HG  and  make  HG—BD.  Draw 
^4£and  at  C,  erect  the  perpen- 
dicular FC,  and  draw  FK.  Then 
KE=FC—EN,  and  KN  is  the  re- 
quired inscribed  square.  For,  in 
the  similar  triangles  AHG  and 
ACF,  we  have  AH:GH::AC  : 
FC,  or  a-\-b:a::b\FC.  By  inver- 
sion, and  then  by  Division,  a\b 
::a—FC:FC,or£I:FC.  In  the 
similar  triangles  ABC  and  KBE, 
AC:KE::BD:BI,  or  BD'ACr. 
BI-KE.  Whence  a:b\\BI:KE.  FIG'  12' 

:.BI\KE\\BI\FC.       :.KE=FC*n&  the   figure   KN  ^  has   its  sides  equal 
and  its  angles  right  angles  by  construction.     Hence,  it  is  a  square.     Q.E.D. 

\Detnonstr  ation.  —  Let  AB=k,  the  chord  of  half  the  arc  ABC^  and  BD 
=a,  the  height  of  the  arc  ABC.     Draw  the   diameter  BE  and  draw  the 


208  FINKEL'S  SOLUTION  BOOK. 

I.     What  is  the  diameter  of  a  circle  of  which  the  height  of 
an  arc  is  5  m.  and  the  chord  of  half  the  arc  10  m.? 

By   formula,  Z>  = 
20  m. 


II. 


1.  10  m.=AB,  the  length  of  chord 

of  half  the  arc. 

2.  5  m.=BD,  the  height  of  arc. 

3.  100  m.2— square  of  chord. 

4.  .-.  20  m.=10Oj-5=BE,  the  diam- 

eter of  the  circle. 


III.     .'.  The  diameter  of  the  circle    is    20 

meters.  FIG  13 

Prob.  XX.     To  find  the  height  of  an  arc,  having-  given  the 
chord  of  the  arc  and  the  radius  of  the  circle. 

Formula. — a—R — *JR*— c*  ,  in  which,  ^?=radius  and  c 
=i  the  chord. 

*Rule. — From  the  radius,  subtract  the  square  root  of  the  dif- 
ference of  the  squares  of  the  radius  and  half  the  chord, 

I.     The  chord  of  an  arc  is  12  feet  and  the  radius  of  the  circle 
is  10  feet.     Find  the  height  of  the  arc. 

By  formula,  a—R — *Jfiz—c  2=10 — VlO2— 62=2  ft. 

1.  10  ft.=the  radius  of  the  circle. 

2.  100  sq.  ft.=square  of  the  radius. 

3.  12  ft.=the  chord. 

4.  6  ft.=half  the  chord. 

5.  36  sq.  ft.=square  of  half  the  chord. 

6.  8  ft.=\/ioo 36  =  square   root   of  the    difference    of  the 

squares  of  the  radius  and  half  the  chord. 

7.  .'.  10  ft — 8  ft.=2  ft.=height  of  the  arc. 

III.     .-.  The  height  of  the  chord  is  2  feet. 

Prob.  XXI.     To  find  the  chord  of  half  the  arc,  having- 
given  the  chord  and  height  of  an  are. 

Formula. — > 


n. 


radius  A  O.  The  triangles  ADB  and  BAE  are  similar,  because  their  an- 
gles are  equal.  Hence,  BE\AB\'.AB:BD,  or  BE:k::k:a.  Whence  BE 
z=Z>r=£2-5-a.  Q.E.D. 

N.  B.—  (1)  If  a  and  D  are  given,  £=y  'Z>X«;    (2)    if    D   and    k    are   given 


*  Demonstration.—  In     Fig.    13,    we    have     BD=BO—DO.      But   Z>O= 
2—£>A2=tf[jR2—c?].     /.   a—R—^R2—c2.      If  a  and  7?  are  given,  (1) 
a)2]  =  2>J(2ajR—  a2);    if  a   and  2c  are    given, 


MENSURATION. 


209 


*Rule. —  Take  the  square  root  of  the  sum  of  the  squares  of  the 
height  of  arc  and  half  the  chord. 

I.     Given   the   chord=48.  the   height=10,  find    the    chord  of 
half  the  arc. 

1.  48=the  chord. 

2.  576=4  of  482=square  of  half  the  chord. 

3.  1Q=  height  of  chord. 

4.  100— vquare  of  height  of  chord. 

5.  676=576+100=sum    of  square    of  half  of  chord   and 

height. 

6.  26=^676— square  ro°t  °f  sum  °f  square   of  half  of  chord 

and  height. 
III.  ^  /.  The  chord  of  half  the  arc  is  26. 

Prob.  XXII.    To  find   the  chord   of  half  an    arc,  having: 
given  the  chord  of  the  arc  and  the  radius  of  the  circle. 


Formula.  —  k=%R*—  ^?\/4/?2—  4c2. 

.  —  Multiply  the  radius  by  the  square  root  of  the  differ- 
ence  of  the  squares  of  twice  the  radius  and  the  chord;  subtract 
this  product  from  twice  the  square  of  the  radius  and  extract  the 
square  root  of  the  difference. 

I.     Given  the  chord  of  an  arc=6  and  the  radius   of  the  circle 
=5,  find  the  chord  of  half  the  arc. 

By 


1.  5=the  radius  of  the  circle. 

2.  10=twice  the  radius  of  the  circle. 

3.  100=square  of  twice  the  radius. 

4.  6=chord  of  the  arc. 

5.  36=square  of  the  chord. 

6.  100  —  36=64=difFerence  of  squares  of  twice   the  radius 

and  the  chord. 

7.  8=\/0-i~scluare  r°ot  of  the  above  difference. 

8.  40=5X8—  the  product  of  the  above  square  root  and  the 

radius. 

9.  50=2x52=twice  the  square  of  the  radius. 

10.  \/oO—  40=V'lO=chord  of  half  the  arc. 
III.      .'.  The  chord  of  half  the  arc  is  \/i(j. 


II. 


*  Demonstration. — In  Fig.  13,  we  have  AB=A 
\/«2-)-c2.     .-.£=V<72-f-c2.  If  £  and  2c  are  given,  (1)#=:^ 

^Demonstration. — From  Prob.  XXI., we  have  &=^az-\-c2. 
XX.  we    have  a=j 


if  k  and  a   are 

From      Prob. 
—  c2.      Substi- 


tuting  this  value  of  az  in  the  above  equation,  £=V2JR2 — 


210 


FINKEL'S  SOLUTION  BOOK. 


Prob.  XXIII.    To  find  the  side  of  a  circuinscribed  polygon, 

he  circle  and  a  side  of  a  simi- 


having- given  the  radius  of  the 
lar  inscribed  polygon. 


.  —  K'-=. 


in  which   K'  is  the  side  of 


V4/?2—  A'2 

the  circumscribed  polygon  and  TTthe  side  of  a  similar 
inscribed  polygon. 

Divide  twice  the  product  of  the  side  of  the  inscribed 
'polygon  and  radius  by  the  square  root  of  the  difference  of  the 
squares  of  twice  the  radius  and  the  side  of  the  inscribed  polygon. 
I.     When  /?=!,  find  one  side  of  a  regular  circumscribed   do- 
decagon. 

^KR  ^K 

By  formula,  K'=  ..  —  .  The   formula   does 

^Ri—K*     *J±—K* 

not  lead  to  a  direct  result,  since  K  is  not  given.  But 
by  the  formula  of  Prob.  XXI.,  if  k  is  replaced  by  K 
we  have  K—Vz  —  V<±—  -1  f°r  2c=l,  since  it  is  the  side 
of  a  regular  inscribed  hexagon,  and  K=V%  —  \/3>since 
2c  is  a  side  of  a  regular  inscribed  dodecagon. 


, 

~ 


VII. 


^ 


RECTIFICATION     OF     PLANE     CURVES     AND 
QUADRATURES  OF  PLANE   SURFACES. 


1.  To  ^Rectify  a   Curve  is  to  find  its  length.     The  term 
arises  from  the  conception  that  a  right  line  is  to  be  found  which 
has  the  same  length. 

2.  The   Quadrature  of  a  surface  is  finding  its  area.    The 
term  arises  from   the   conception   that   we   find   a  square  whose 
area  is  equal  to  the  area  of  the  required  surface. 

The  formula  for  the  rectification  of  plane  curves  is 

dx,  when   the   curve    is   re- 


ferred to  rectangular  co-ordinates. 


curve  is  referred  to  polar  co-ordinates. 


*     2dz    2 


are  formulae  for  the  rectifica- 
tion of  curves  of  double  cur- 
vature, when  referred  to  rec- 
tangular co-ordinates*. 


MENSURATION.  211 


are  formulae  for  the 
rectification  of 
j     curves    of    double 
'r   curvature,  referred 
to    polar    co-ordi- 
nates. 

A=  Cydx  or    Cxdy  is   the   formula  for  the  quadrature  of  any 
plane  surface  referred  to  rectangular  co-ordinates. 

A.=?  C^rzdO  is  the  formula  for   the   quadrature  of  plane  surfaces, 
referred  to  polar  co-ordinates. 

3.  A  Surface  Of  Revolution  is  the  surface  generated 
by  a  line  (right  or  curved)  revolving  around  a  fixed  right  line  as 
an  axis,  so  that  sections  of  the  volume  generated,  made  by  a  plane 
perpendicular  to  the  axis  are  circles. 


/  y  ->J  !  1-f-  1  —        {  dx  is  the  formula  for  a  surface  of  revo- 
JJ  N  r      '   \dx\     ) 

lution,  referred  to  rectangular  co-ordinates. 
=27T  fyds=&7t  Cr  sin  6\     r*+\-^\        dQ  is  the  formula  for 


a  surface  of  revolution,  referred  to  polar  co-ordinates. 

K=TT  fy*dx  or  xzdy  is  the  formula  for  the  volume  of  a  solid 
of  revolution  referred  to  rectangular  co-ordinates. 

V=  C  C  Cdxdydz  and  V=  C  Czdxdy  are  formulae  for  the 
cubature  of  solids,  requiring  triple  and  double  integration. 

F=  f  CzrdBdr  and  V=CJJr2  sin  SdcpdQdr  are  the  formulas 
for  cubature  of  solids  referred  to  polar  co-ordinates.  From 
the  equation  to  the  surface  of  the  solid,  z  must  be  expressed 
as  a  function  of  r  and  6. 

__j^>2  is  the  rectangular  equation   of  a  circle   referred  to 
the  center. 

y2==2-/?#  —  xz  is  the  rectangular  equation  of  a  circle  referred  to 
the  left  hand  vertex  as  origin  of  co-ordinates. 

r=%R  cos.#  is  the  equation  of  the  circle  referred  to  polar  co-ordi- 
nates. 

Prob.  XXIV.    To  find  the  circumference  of  a   circle,  the 
radius  being  given. 


212  FINKEL'S  SOLUTION  BOOK. 


"    —Rdx  1        1.3       1.3.5 


=4ff  Xl-570796+=3.141592x2.ff=2;rJff,  in  which 

z..o.o. 

7T=3.141592+.     Since  the  diameter  is  twice  the  radius,  we  have 
27rft=7tD,  in   which    D  is    the    diameter.      .*.   C,=%7tR=7tD. 

C  C 

.'.  (1)  R=—^,  (2)Z>=  —  ,  where  C  is  the  circumference. 

£  TC  7i 

Rule.  —  Multiply  twice  the  radius  or  the  diameter  by  3.141592. 

I.      What  is  the  circumference  of   a  circle  whose  radius  is  17 
rods  ? 

By    formula,    C=2^/?=3.141592  X  34  rods  =  106.814128 
rods. 

1.  17  rods=the  radius. 

2.  34  rods=2Xl7  rods=the  diameter. 

3.  106.814128  rods=3.141592X34rods=the  circumference. 

III.     .-.  The  circumference  is  106.814128  rods. 

Note.  —  The  ratio  of  the  circumference  to  the  diameter  can  not  be  exactly 
ascertained.  An  untold  amount  of  mental  energy  has  been  expended  upon 
this  problem;  but  all  attempts  to  find  an  exact  ratio  have  ended  in  utter 
failure.  Many  persons  not  noted  along  any  other  line,  claimed  to  have 
found  this  clarem  impossibilitibus  by  which  they  have  unlocked  pll  the  diffi- 
culties that  have  encumbered  the  quadrature  of  the  circle  for  more  than  two 
thousand  years.  The  Quadrature  of  the  Circle  is  to  find  a  square  whose  area 
shall  be  exactly  equal  to  that  of  the  circle.  This  can  not  be  done,  since  the 
ratio  of  the  circumference  to  the  diameter  can  not  be  exactly  ascertained. 
Persons  claiming  to  have  held  communion  with  the  "gods"  and  extorted 
from  them  the  exact  ratio  are  ranked  by  mathematicians  in  the  same  class 
with  the  inventors  of  Perpetual  Motion  and  the  discoverers  of  the  Elixir  of 
Life,  Alkahest,  the  Fountain  of  Perpetual  Youth,  and  the  Philosopher's 
Stone.  Lambert,  an  Alsacian  mathematician,  proved,  in  1761,  that  this  ratio 
is  incommensurable.  In  1881,  Lindemann,  a  German  mathematician,  dem- 
onstrated that  this  ratio  is  transcendental,  and  that  the  quadrature  of  the 
circle  by  means  of  the  ruler  and  compass  only,  ort>y  means  of  any  algebraic 
curve,  is  impossible.  Its  value  has  been  computed  to  several  hundred  deci- 
mal places.  Archimedes,  in  287  B.  C.,  found  it  to  be  between  3f?  and  3f  ; 
Metius,  in  1640,  gave  a  nearer  approximation  in  the  fraction  ff  |;  and,  in  1873, 
Mr.  W.  Shank  presented  to  the  Roj^al  Society  of  London  a  computation  ex- 
tending the  decimal  to  707  places.  The  following  is  its  value  to  600  deci- 
mal places: 

3.  141,  592,  653,  589,  793,238,462,643,383,  279,  502,884,  197,  169, 
399,375,105,820,974,944,592,307,816,406,286,208,998,628,034,825, 
342,117,067,982.148,086,513,282,306,647,093,844,609,550,582,231, 
725,359,408,128.481,117,450-284,102.701,938,521,105,559,644,622, 
948  ,954,930,381  ,964,428,810,975  ,665,933,446,  128  ,475  ,648  ,233,786, 


MENSURATION.  213 

783,165,271,201,909,145,648,566,923,460,348,610,454,326,648,213, 
393,607,260,249,141,273,724,587,006,606,315,588,174,881,520,920, 
962,829,254,091,715,364,367,892,590,360,011,330530,548,820,466, 
521,384,146,951,941,511,609,433,057,270,365,759,591,953,092,186, 
117,381,932,611,793,105,118,548,074,462,379,834,749,567,351,885, 
752,724,891,227,938,183,011,949,129,833.673,362,441,936,643,086, 
021,395,016,092,448,077,230,943,628,553,096,620,275,569,397,986, 
950,222,474,996,206,074,970,304,123,669+. 


Bernoulli's  Formula. 


"''  Wallis's  Formula>  1666. 


i7T=l+l_.  .......  Sylvester's  Formula,  1869. 

1+1.2 
1+2.3^ 
1+3.4 

M-L 

1+4.5 


2+  ......  Buckner's  Formula. 

The  Greek  letter  n,  was  first  used  by  Euler,  to  designate  the 
ratio  of  the  circumference  to  the  diameter. 

Prob.  XXV.  To  find  the  length  of  any  arc  of  a  circle, 
having-  given  the  chord  of  the  arc  and  the  height  of  the  arc, 
i.  e.,  the  versed  sine  of  half  the  arc. 


(a). 

Rdx 


2)3 
2  \3    ' 


the  arc  and  c—  half  the  chord  of  the  arc. 


*Note.  —  This  series  was  discovered  by  Bernoulli,  but  he  acknowledged 
his  inability  to  sum  it.  Euler  found  the  result  to  be  ^TT*.  For  an  inter- 
esting discussion  of  the  various  formulae  for  TT,  see  Squaring-  the  Circle, 
Britannica  Encyclopedia 


214  FINKEL'S  SOLUTION  BOOK. 

(b.)  Formula. — s=a.rc=^(8& — 0)*,  where  a  is  the  chord  of 
the  whole  arc  and  b  the  chord  of  half  the  arc. 

Rule  from  (b) .  From  eight  times  the  chord  of  half  the  arc  sub- 
tract the  chord  of  the  whole  arc ;  one-third  of  the  remainder  will 
be  the  length  of  the  arc,  approximately. 

I.     Find  the  length  of  the  arc  whose  chord  is  517638  feet 
and  whose  half  chord  is  261053.6  feet. 

By  formula  (b),  s=%(8&—a)=%  (8X261053.6—517638)= 
52359.88  feet. 

f  1.     261053.6  feet=length  of  chord  of  half  arc. 

2.  2088428.8  feet=8X261053.6  feet=eight  times  the 

length  of  chord  of  half  arc. 

3.  517638  feet=rlength  of  chord  of  whole  arc. 


II.  < 


4.     1570790.8  feet=2088428.8  feet— 517638  feet=differ- 


ence  between  eight  times  chord  of  half  arc  and 
chord  of  whole  arc. 

5.     52359.69  feet=J  of  1570790.8  feet=length  of  arc, 
nearly. 

III.     .-.  The  length  of  the  arc  is  52359.69  feet. 

Note. — This  important  approximation  is  due  to  Huygens,  ( he  wrote  his 
name  Hugens.  It  is  also  sometimes  spelled  Huyghens),  a  Danish  mathe- 
matician, born  at  the  Hague,  April  14,  1629,  and  died  in  the  same  town  in 
1695.  For  a  brief  biography  of  this  noted  mathematician,  see  Ball's 
A  Short  History  of  Mathematics,  pp.  302-306. 

The  following  is  Newton's  demonstration  : 

Let  R  be  the  radius  of  the  circle,  L  the  length  of  the  arc,  A  the  chord 
of  the  arc,  and  B  the  chord  of  half  the  arc. 

A  L       B  L 

Then  -^=2  sm.^,   -^=2  sm.^. 

X         XZ        X*> 

Since,  sin.  #=——- gj-j-yj-— etc.  (See  Bowser's  Treatise  on  Trigonom- 
etry,  or  any  other  good  work  on  the  subject),  we  have 

and 


4^=2 (  2*-V2.#)   +V2^/  -etc.    ) 

*      V~    ~~3T~     ~~5T~          / 

£=<?- 


.  * .  A — L     A?  £>2  +  o4^i  £>4     etc.,  and 
8#=4Z,— 


,  nearly,=A  nearly. 

In  the  problem  proposed,  the  radius  is  100000  feet  and  the  arc  is  30°. 
Using  7r=3 . 1415926,  £=52359 . 88  feet.  .  • .  The  result  by  the  formula  lacks 
only  about  2  inches  of  being  the  same. 


MENSURATION.  215- 


(c.)    *orm^-^=*rc=a(  x 

This  formula  is  a  very  close  approximation  to  the  true  length 
of  the  arc  when  a  and  c  are  small.  The  first  formula  may  be  ex- 
tended to  any  desired  degree  of  accuracy. 

Rule  from  (c).  —  Divide  10  times  the  square  of  the  height  of 
the  arc  by  15  times  the  square  of  the  chord  and  33  times  the  height 
of  the  chord;  multiply  this  quotient  increased  by  1,  by  2  times  the 
square  root  of  the  sum  of  the  squares  of  the  height  and  half  the 
chord. 

I.  The  chord  of  an  arc  is  25,  and  versed-sine  15,  required  the 
length  of  the  arc. 

02_|_cBr-c2__02  (C2_a2}3         3 

By   formula  (a),  arc=  -±-  [^-^  [^^  \    +^ 

5    (c*—a*Y     &    -]_152-t-252r252-152     1 
l2|/ia+c2J   '      'CJ~     2X15    Ll52+252~i~6X 

3  r252-152]5       5   r26»-15n  7          -] 
152+252J    ^   152+252J    +n  &C 


+ft. 

1.  25  ft.=length  of  the  chord. 

2.  15  ft.— height  of  the  arc,  or  the  versed-sine. 

3.  2250  sq.  ft.=10  times  152=10  times  the  square  of  the 

height  of  the  arc.  [chord. 

4.  9375  sq.  ft.=15  times  252=15  times  the  square  of  the 

5.  7425  sq.  ft=33  times  152=33  times  the  square  of  the 

height  of  arc. 
!!.<{    6.  17HOO  sq.  ft.=7425  sq.  ft.+9375  sq.  ft. 

7.  ^=^2250-7-17800=10  times  152-r-(15  times  252+33 

times  15s). 

8.  14^==4££==1  +  10  times  152-j-(15  times  252+33 

times  152). 

9.  381J  sq.  ft.=152+(12j)2. 

10.  53.58   ft.=-|fixVl52+(12i)2=|fiX|V61^=length  of 

arc,  nearly. 
III.     .'.  53.58  ft.=length  of  the  arc. 

Prob.  XXVI.    To  find  the  area  of  a  circle  having  given  the 
radius,  diameter,  or  circumference. 


Formula.— 


C,  when  the  ra- 


216  FINKEL'S  SOLUTION  BOOK. 

dius   and    circumference  are  given.     /.  (1)  R=*jA-r-rt,  (2)  D= 
=2  R=2V.4-v-;r,  and  (3)  C=\/4^4= 


Rule  I.  —  The  area  of  a  circle  equals  the  square  of  the  radius 
multiplied  by  3.14-1592;  or  (2)  the  square  of  the  diameter  multi- 
plied by  .785398;  or  (3)  the  square  of  the  circumference  multiplied 
by  .07958/  or  (4)  the  circumference  multiplied  by  \  of  the  diame- 
ter; or  (5)  the  circumference  multiplied  by  \  of  the  radius. 

Rule  II.  —  Having  given  the  area.  (1)  To  find  the  radius: 
Divide  the  area  by  3.14159%,  and  extract  the  square  root  of  the 
quotient.  (2)  To  find  the  diameter  :  Divide  trie  area  by  3.141592 
and  multiply  the  square  root  of  the  quotient  by  2.  (3)  To  find 
the  circumference:  Multiply  the  area  by  3.141592  and  multiply 
the  square  root  of  the  product  by  2. 

I.     What  is  the  area  of  a  circle  whose  radius  is  7  feet? 

By  formula,  ^=7T.ff2=3.141592x72=153.93804+  sq.  ft. 


1.  7  ft.=the  radius. 
5.  49  sq.  ft.=72=square  of  the  radius. 

153.93804  sq.  ft.=3. 141592x49  sq.  ft.=area  of  the  circle. 

III.     .'.  153.93804  sq.  ft.=area  of  the  circle. 


IL/2. 
13. 


I.     What  is  the  area  of  a  circle  whose  diameter  is  4  rods  ? 

By  formula,  ^=i7rZ>2=4x3.141592x42=12.566368  sq.  ft. 

II.  4  ft.=the  diameter. 
2.  16  sq.  ft.=square  of  the  diameter. 
3.  12.566368  sq.  ft.=ix3.141592x42=.785398Xl6  sq.  ft. 
=area  of  the  circle. 

III.     .-.  12.566368  sq.  ft.^area  of  the  circle. 
I.  What  is  the  area  of  a  circle  whose  circumference  is  5  meters  ? 


By  formula,  ^4=.=^=1.989  m2. 

{1.  5  m.=the  circumference. 
2.  25  m.2=the  square  of  the  circumference. 
3.  1.989  m.2=.07958x25  m.2=the  area  of  the  circle. 
III.     /.  1.989  m.2=the  area  of  the  circle. 

Remark.  —  We  might  have  found  the  radius  by  formula  (1)  under  Prob. 
XXIV  and  then  applied  the  first  of  Rule  I.  above.  We  might  have  found  the 
radius  by  formula  (1)  of  Prob.  XXIV  and  then  applied  (5)  of  Rule  I.  above. 

I.     What  is  the  circumference  of  a  circle  whose  area  is  10  A.? 


MENSURATION. 


217 


By  formula  (3),  C=2*/7rA=M3. 141592 X  1600=80V?r80X 
1.7724539—141.796312  rods. 

1.  10  A.—1600  sq.  rds.=the  area  of  the  circle. 

2.  1600H-7T=the  square  of  the  radius. 

II.< 


3.  .-.  40        —V^the  radius. 


71 


80  ,_  40  - 

~7t  ^7f==7r^  times  ~V7T=the  diameter. 

40 


5.  SQ\x=7TX  —VTT—  141.796312  rods=the  circumference. 
III.     .'.  141.796312  rods=the  circumference  of  the  circle. 

I.     With  what  length  of  rope  must  a  horse  be  tied  to  a  stake 
so  that  he  can  graze  over  one  acre  of  grass  and  no  more? 

By  formula  (1),  7?=Vyl~^=Vl60-r-7r=4>|—  =7.1364+  rd. 

1.   1  A.=160  sq.   rd.=area  of   the   circle   over   which  the 

horse  can  graze. 
IL)2.   160-7-  7r==square  of  the  radius. 

3.  Vl60-r-7T=4VlO-7~7T=7.1364  rd.=radius  or  length  of  rope. 
III.     .-.  7.1364  rd.=length  of  the  rope. 

Prob.  XXVII.  To  find  the  area  of  a  Sector,  or  that  part  of 
a  circle  which  is  hounded  by  any  two  radii  and  their  included 
arc,  having  griven  the  chord  of  the  arc  and  the  height  of  the 
arc. 


-      i*       cz—a*  (  (az+c*}  2      (c*—  a*}*} 

sm  7r-^l^  ~        t 


+ 

:.j  in  which  c  is  half  the  chord  of 
arc  and  a  the  height  of  arc. 

Demonstration. — Let  AB=x,  BD=y,  and  ft=A£>=thc  radius  of  the  cir- 
cle. Then  x2-}-y2=K2,  the  equation  of  the  circle  referred  to  the  center. 
Now  A— 2  Cydx;  buty=(7?2 — x2)lA,  from  the  equation  of  the  circle. 

/.  A=.2  C(R2—x*}Xdx=x  (R2—x2}X-\-R*   sin~1-^.  But  x=R— a  and  y=c. 


Hence  ^=(7?— a)[R*— (/?— a)2]+7?2  sin~~.  Ikit,from(2)  Prob.  XX, 


218 


FINKEL'S  SOLUTION  BOOK. 


Rule.—  (1)  Find  the  length  of  the  arc  by  Problem  XXV, 
and  then  multiply  the  arc  by  half  the  radius  which  may  be  found 
by  Problem  XX,  in  'which  c  and  a  are  known  and  R  is  the  un- 
known quantity. 

(2)  If  the  arc  is  given  in  degrees,  take  such  a  part  of  the 
whole  area  of  the  circle  as  the  number  of  degrees  in  the  arc  is 
ef360*. 

I.  Find  the  area  of  the  sector,  the  chord  of  whose  arc  is  40 
feet,  and  the  versed-sine  of  half  the  arc  15  feet. 


By  formula, 


202— 


c  202—  152 


1       202—15 


II. 


III. 


202+152 
1.  53.58   ft.= 


,3  ,        1  _  f20»—152'|6 
J   +1.2.3.4.5  l20»+15»J    " 


length  of  the  arc,  by  (£),  Prob.  XXV. 


2.  20f  ft.= 


#2    I    C2 

— 


=radius  of   the   circle,  by   solving    th 


formula  of  Prob.  XX  with  respect  to  R. 
3.  .'.  558.125  sq.  ft.=-J  (20-|x53.58)=area  of  the  sector. 

.'.  558.125  sq.  ft.=the  area  of  the  sector. 


Trigonometry,  we   have  sin~lO=0 — -^ 


But'fron 


_#s — £Ci      Hence,  A=z 


2a 


In  this  formula, 


c2—a*)  *  c  r2—  a2  __  1_  [  r2—  a*    3  .         1        (c2—  a^  <t  ^ 

2a    }     i  c^+a*     1.2.3  U2+«2J       1-2.3.4.5  U2+«2J   "         \ 


c  is  the  area  of  the  tri- 


angle   DBA.     For  *:=/?—  «= 


^a-LcZ  c*—a* 

-JT a— n : 

2a  2a 


the  altitude  and  c  is  half  the  base  of  the  triangle 

(C2  _  a2} 

.:    —  -  -  \c  =  the   area  of   the  triangle   DBA. 

Therefore,  if  we  subtract  the  area  of  the  tri- 
angle DBA  from  the  area  of  the  sector,,  we 
shall  have  the  area  of  the  segment  DEC.  Hence, 

(cz  —  az\  2 

-the    area  of    the    segment    DEC   is    —  -  — 

2a 


be  carried  to  any  desired  degree  of  accuracy. 


FIG    74. 


MENSURATION.  219 

I.     What  is  the  area  of  a  sector  \vhose  arc  is40°and  the  radius 
of  the  circle  9  feet? 

1.  9  ft.=radius  of  the  circle. 

2.  7n??2=7z92=area  of  the  whole  circle. 
IU3.  40°=length  of  the  arc. 

4.  40°Hrof360°. 

5.  7r9=4  of  7r92=28.274328  sq.  ft.=area  of  the  sector. 
III.  '  .-.  The  area  of  the  sector  is  28.274328  sq.  ft. 

Prob.  XXVIII.  To  find  the  area  of  the  segment  of  a  circle, 
having-  given  the  chord  of  the  arc  and  the  height  of  the 
segment,  i.  e.,  the  versed-sine  of  half  the  arc. 


(cz_az}         1     (c*—  a*} 

__ 


, 

-&c- 


.  —  Divide  the  cube  of  the  height  by  twice  the  base  and  in- 
crease the  quotient  by  two-thirds  of  the  -product  of  the  height  and 
base. 

I.     What  is  the  area  of  a  segment  whose  base  is  2  feet  and  al- 
titude 1  foot? 

By  formula  (*),  A^ 


* Demonstration. —  In  the  last  figure,  let  jB  C=a=altitude  of  the  segment  and 
DE—2c=\hQ  base  of  the  segment.    Then  BDZ= BCXSF=a  (2R— a)—cz. 

Whence  R=C  J~rt  ,  and  AD=R — a=C    '  a  — «~ C       a  .   Z?C=£— v/ZM^rz^ 
2a  2a  2« 

/?  77  r 

By   Trigonometry,—— r=sin/Z>^4C,  or—  rrr sin/ DAC.      Now   27r^?r= 
yl  .Zx  oi 


180°  ,    180°  . 

--    Therefore,  R:  arcDC::  -  :  ?=  -  -  —  X  --          Let   s  =  arc 

7T  7T  R  7T 


DCE.  Then  the  /DAC=-=.     .'.     --^sin--—  .     In  like  manner,  from 

ft        Zff  JTL  Zfi 

the  right  angled  triangle  ^Z>C,-,=rsin/C^Z?,  or  since  the 


^  /  CAD,  ^5— sin— .       Now  since  the  sine  of  any  angle  fczfJ —  — — 
.,  the  above  equation  becomes 
— &c (1),  and 


ff 


— &c (2).     Multiplying  equation  (2) 

by    8  and  subtract  equation  (1)  in  order  to   eliminate  the  term  containing 
s*,  we  have  approximately ,^^-C=r^|— %  fr-^-^Tl  (nTs)   +  &c.  Omitting 


the  negative  quantity,  since  it  is  very  small  in  comparison  with  s  and  be- 
cause it  is  still    more  diminished  by  a  succeeding  positive  quantity,  we  have 


II. 


FINKEL'S  SOLUTION  BOOK. 

1.  1  ft.=altitude  of  the  segment. 

2.  2  ft.=base  of  the  segment. 

3.  4  ft.=twice  the  base  of  the  segment, 

4.  1  cu.  ft— cube  of  the  height  of  the  segment. 

5.  i  sq.  ft.=l-7-4=quotient  of  the  cube  of  the  height  and 


twice  the  base. 

6.  2  sq.  ft.=2  X  l=product  of  the  height  and  base. 

7.  lij-  sq.  ft.=f-  of  the  product  of  the  height  and  base. 

1-J-  sq.  ft.-f-^  sq.  ft.=lT7^sq.  ft.==area  of  the  segment. 
III.     .•.  The  area  of  the  segment  is  1T7^-  sq.  ft. 

Prob.  XXIX.  To  find  the  area  of  a  circular  zone,  or  the 
space  included  between  any  two  parallel  chords  and  their 
intercepted  arcs. 

,    \A        c*—a2*c*—a2         1 
.-a  A= 


)      __  (c'*—a 

"        '  1  "          M 


1.2.3 


Rule.  —  Find  the  area  of  each  segment  by  Prob.  XXVIII.,  and 
take  the  difference  between  them,  if  both  chords  are  on  the  same 
side  of  the  center;  if  on  opposite  sides  of  the  center,  subtract  the 
sum  of  the  areas  of  the  segments  from  the  'whole  area  of  the  circle^ 

I.      What  is  the  area  of  a  zone,   one  side   of  which  is  96,  and 
the  other  60,  and  the  distance  between  them  25? 

Let  ^4^=60=2^,  CY?=96=2c,  and  HK=2b=h.  Then  AH 
=30=^  and  CK=±S=c.  Let  OA=R.  Then 

But 


ne.  is. 


—  3  —  —  —   -  g        —  =%(4\/c2+a2—  a).        This    is    the    approximate 

length  of  an  arc  in  terms  of  its  height  and  base.      Now  the  area  of  the  seg- 
ment £>CJS=%  A  CXarcDCE—  area  of  the  triangle  DEA—yz  RxS—%  AB 

X 


Q.  E.  D. 


MENSURATION.  221 


—  h*  —  c/'a')s.      In  like  manner,  LK—a=R  —  i/J?z  _  c*,=i 


.-.  By  formula  («),  ^= 


X 


(488—  252—  3Q2)=radius  of  the  circle. 

2.  OK=>lR*—c*  =V503— 48g=14. 

3.  .*.  XA==«==50 — 14=36— altitude  of  segment 

4.  O  H — v ' R  ^ c'  2 " — V^5Q 2 30  *  — 40. 

5.  .-.  Z^T=«/=50 — 40=10=altitude  of  the  segment 

363 

+f(96X36)=2547=area  of  segment   CDBLA. 


103 

7.  x — —  4-f(60X10)=408i=area  of  the  segment 

8.  .'.2547 — 408i==2138f=area  of  the  zone   CDBA. 
III.     .'.  2138f=area  of  the  zone  ABDC. 

Note. — This  result  is  only  approximately  correct.  The  radius  of  the  cir- 
cle may  be  found  by  the  following  rule: 

Subtract  half  the  difference  between  the  two  half  chords  from  the  greater 
half -chord,  multiply  the  remainder  by  said  difference,  divide  the  product  by 
the  width  of  the  zone,  and  add  the  quotient  to  half  the  width.  To  the  square 
of  this  sum  add  the  square  of  the  less  half  chord,  and  take  the  square  root  of 
the  sum. 

This  rule  is  derived  from  the  formula  in  the  above  solution,  in  which 


222  FINKEL'S  SOLUTION  BOOK. 

Prob.  XXX.  To  find  the  area  of  a  circular  ring,  or  the 
space  included  between  the  circumference  of  two  concen- 
tric circles. 

Formula,* — («.)  A=n  (7?2 — r2),  in  which  R  and  r  are 
the  radii  of  the  circles. 

(b.)  *A=%7rc*,  in  which  c  is  a  chord  of  the  larger  circle 
tangent  to  the  smaller  circle. 

I.  Required  the  area  of  a  ring  the  radii  of  whose  bounding 
circles  are  9  and  7  respectively. 

By    formula    («),    A  =  7t(R*  —  r*)  =  7r(92—  72)=32?r= 

100.530944. 
'1.  9=/?=:radius  of  the  larger  circle,  and 

2.  7=7'=radius  of  the  smaller  circle. 

3.  7r92=7r^?3=area  of  larger  circle,  and 

4.  7r72=7T?'2=area  of  smaller  circle. 

5.  .'.  7T93  —  7r72  =  7r(9s—  72)  =  327T= 

100.530944=area  of  the  ring. 
III.     /.  100.530944=the  area  of  the  ring. 

* Demonstration. — Let  ABC  be  the  chord  of  the 
large  circle,  which  is  tangent  to  the  smaller 
circle,  and  let  A£C=c.  Then  BC=%c= 


II.< 


and  Jf7rc2r=/r(^?2_r2)      But  K(R*—  r2)  is  the  dif- 

ference of  the  areas  of  the  two  circles  or  the  area 

of    the  ring,      *.  ^7rc2-^:the   area  of  the   ring. 

Q    £  £).  FIG-  16' 

Prob.  XXXI.    To  find  the  areas  of  circular  lunes,  or  the 
spaces  between  the  intersecting*  arcs  of  two  eccentric  circles. 


Formula.—  ^= 


*  —  Find  the  area  of  the  two  segments  of  'which  the  lunes 
are  formed^  and  their  difference  'will  be  the  area  required. 

I.     The  chord  AB  is  20,  and  the  height  DC  is  10,  and  DE  2; 
find  the  area  of  the  lune  AEB  C. 


+</*.  V  . 

If  now  we  find  the  altitudes  of  the  two  segments  and  then  find  the  length 
of  the  arcs  of  the  segments  by  formula  (£),  Prob.  XXV,  and  then  find  the 
area  of  the  sectors  by  multiplying  the  length  of  the  arcs  by  half  the  radius, 
from  the  areas  of  the  sectors  subtract  the  triangles  formed  by  the  radii  of 
the  circles  and  the  chord  of  the  arcs,  we  shall  then  have  the  area  of  the  two 
segments.  Taking  their  difference,  we  shall  have  for  the  area  of  the  zone 
2136*75,  which  is  a  nearer  approximation  to  the  true  area. 


MENSURATION. 


223 


By  formula,  A= 


II. 


2.  DE  =  height     of    segment 

3.  DC=  height     of     segment 

area  of  the  segment  A  CBD. 


FIG.  n. 

area  of  the  segment  AEBD. 
III.     .-.  158^  —  26ff=131TV=area  of  the  lime  A  CBE. 


VIII.     CONIC  SECTIONS. 


1.  The   ConiG  Sections  are    such    plane    figures    as    are 
formed  by  the  cutting  of  a  cone. 

2.  If  a  cone  be  cut  through  the  vertex,  by  a  plane  which  also 
cuts  the  base,  the  sections  will  be  a  triangle. 

3.  If  a  right  cone  be  cut  in  two  parts,  by  a  plane  parallel  to 
the  base,  the  section  will  be  a  circle. 

4.  If  a  cone  be  cut  by  a  plane  which   passes   through   its  two 
slant  sides  in  an  oblique  direction,  the  section  will  be  an  ellipse. 

5.  The  Transverse  Amis  of  an  ellipse  is  its   longest 
diameter. 

6.  The    Conjugate    Amis   of  an   ellipse    is    its   shortest 
diameter. 

7.  An   Ordinate  is  a  right  line  drawn  from  any  point  of 
the  curve,  perpendicular  to  either  of  the  diameters. 

8.  An  Abscissa  is  that  part  of  the  diameter  which  is  con- 
tained between  the  vertex  and  the  ordinate. 

9.  A   Parabola  is   a   section    formed    by  passing  a  plane 
through  a  cone  parallel  to  either  of  its  slant  sides. 

1C.     The  Axis  of  a  parabola  is  a  right  line  drawn  from  the 
vertex,  so  as  to  divide  the  figure  into  two  equal  parts. 

11.     Tile   Ordinate  is  a  right  line  drawn   from  any  point 
in  the  curve  perpendicular  to  the  axis. 


224  FINKEL'S  SOLUTION  BOOK. 

12.  The  Abscissa  is  that  part  of  the   axis  which  is  con- 
tained between  the  vertex  and  the  ordinate. 

13.  An  Hyperbola  is   a    section    formed   by   passing   a 
plane  through  a  cone  in  a  direction  to  make  an  angle  at  the  base 
greater  than    that  made   by  the  slant  height.     It  will    thus   pass 
through  the  symmetrical  opposite  cone. 

14.  The  Transverse  Diameter  of  an  hyperbola,  is  that 
part  of  the  axis  intercepted  between  the  two  opposite  cones. 

15.  The  Conjugate  Diameter  is  a  line  drawn  through 
the  center  perpendicular  to  the  transverse  diameter 

16.  An   Ordinate  is  a  line  drawn   from   any  point  in  the 
curve  perpendicular  to  the  axis. 

17.  The  Abscissa  is  the  part  of  the  axis  intercepted  be- 
tween that  ordinate  and  the  vertex. 

1.  ELLIPSE. 

a2y2-^-&2x2=a2l>2  is  the  equation  of  an  ellipse  referred  to  the 
center. 

32 
y2=—  (%ax — AT2)  is  the  equation  of  the  ellipse    referred  to  left 

hand  vertex. 

In  these  equations,  a  is  the  semi-transverse  diameter  and  b  the 
semi-conjugate  diameter  \y  is  any  ordinate  and  x  is  the  corres- 
ponding abscissa.  When  any  three  of  these  quantities  are  given 
the  fourth  may  be  found  by  solving  either  of  the  above  equations 
with  reference  to  the  required  quantity. 

p=^— £    is  the  polar  equation  referred  to  the  centre,  and 

1 — e  cos  6 

at  i ez  \ 

p=— -r n    is  the  polar  equation  referred   to    the    left  hand 

l-\-e  cos  6 

vertex. 

Prob.  XXXII.  To  find  the  circumference  of  an  ellipse,  the 
transverse  and  conjug-ate  diameters  being-  known. 


Formula.— cir.=  C= 


MENSURATION. 


1-3  -  y-\        3*6      (502r3a2f02  .  ,tx    *Ari  -  51 
/r2  _  v2    I  __  -  J  -  1  -  I  —  sin     --  \a2  —  x2  \  _ 
*  J     2.4.6«5    (   6   L   2     [2  a     2*  } 

a      Afna      e*  fa    na\ 
=4  VT—A2  '  T)  - 


4     V2'2  ~  * 


r         ^2 
|  !___ 


Rule.  —  Multiply  the  square  root  of  half  the  sum  of  the  squares 
of  the  two  diameters  by  8.141592,  and  the  product  will  be  the  cir- 
cumference, nearly. 

I.  What  is  the  circumference  of  an  ellipse  whose  axes  are  24 
and  18  feet  respectively  ? 


By  formula,   Ctr.=  C=27TX  12 

-JV—  &c.  =27rx  12  X-  87947=66.31056  ft.,  nearly. 


1.  576  sq.  ft.=242=square  of  the  transverse  diameter. 

2.  324  sq.  ft.=182=square  of  the  conjugate  diameter. 

3.  900  sq.  ft.— sum  of  the  squares  of  the  diameters. 

4.  450  sq.  ft.=half  the  sum  of  the  squares  of  the  diameters. 
'  5.  15V2ft.=V450=square  root  of  half  the  sum  of  the  squares 

of  the  diameters. 

6.  7rl5\/2  ft.=  66.6434  ft.,    nearly,  =the    circumference  of 
the  ellipse. 

III.  .-.  The  circumference  of  the  ellipse  is  66.6434  ft.  nearly,  by 
the  rule. 

Prob.  XXXIII.  To  find  the  length  of  any  arc  of  an  ellipse, 
having  given  the  ordinate,  abscissa,  and  either  of  the  diam- 
eters. 

Formula.— s-=Z[%7ta\l— (|)2^ (|-i)24  —  (i-M)2 

J.  o 

<?6  )  — i  .x       e*  fa*  .        x      x  }         e4     f~3a2 

&c  I     I  —  a  sin  1 —     — sin ^a — x2     — — - — -\   — T— 

5  j  J  a      2a[2  a      V  }       2.4«3  L  4 

C  *  2  v  y      . V  3      ^ ,  ^-^     J 

—  sin"1 Wa2 — x2  — -~\a2 — x*        — &c.,  in  which  x  is  the  ab- 

[2  a     2  4  _JJ 

\a*—b* 

scissa:  a  the  semi-transverse  diameter  ;  and  £=->J =— =the    ec- 

a 

ccntricity  of  the  ellipse. 


226  FJNKEL'S  SOLUTION  BOOK. 

Rule.—  Find  the  length  of  the  quadrant  CB  by  Prob.  XXXI 
and  CF  by  substituting  the  value  of  x  in  the  above  series.  Twice 
the  difference  between  these  arcs  will  give  the  length  of  the  arc 
FBG. 

I.  What  is  thelengthofthearc7^G,if  OE=x=$,EF=y=%, 
and  0C=3=10? 

Since  02jx2+32*2=#2£2,  we  find,  by  substituting  the  values  of 
x,y,  and  b,  0=15.  Then  by  the  formula,  FBG=s=&  %na  j  1  — 


x       e*(az          tx      *r—  -  -) 

—  a  sm-1-—  —  —  sin"1  --  7Vaz—x2  \ 

a      2a  (2  a      2  J 


FIG.  18. 

/io\av'  /  i     o     £  \  O    ""  O  /  C\   t  *  **        •  +       u 

—  (I4)2  ^—  (M'f)    K &c*  (  — 2)15sm -1— 


__,—!  / 

-9^J  -  &c.  J  =,rl5x.815-2    j  g* 


.152  f!52  .  _.  9 
IB 


Prob.  XXXIV.    To  find  the  area  of  an  ellipse,  the  trans- 
verse and  conjugate  diameters  being-  given. 


Formula.—  A=4<tx=4 


«2   —x*)dx=nab,  in 

which  a  and  b  are  the  semi-transverse,  and  semi-conjugate  diame- 
ters. 


.  —  Multiply   the  product   of  the    semi  -diameters   by  n=^ 
3.141592,  or  multiply  the  product  of  the  diameters  by  \7t=.785398. 
I.  What  is  the  area  of  an  ellipse  whose  traverse  diameter  is  70 
feet  and  conjugate  diameter  50  feet? 

By  formula,  ^=^^=^35X25=2748.893  sq.  ft. 

rl.  35  ft=|  of  70  ft=length  of  the  semi-transverse  diameter. 
II.<  2.  25  ft=-^  of  50  ft.=length  of  the  semi-conjugate  diameter.. 

13.  /.  2748.893  sq.  ft.=7rx35x25=the  area  ofthe  ellipse. 
III.     .-.  The  area  of  the  ellipse  is  2748.893  sq.  ft. 


NOTE.  —  7ra3  =  ^7r«2.7r^2.     .•.     The  area  of  an  ellipse   is   a   mean  propor- 
tional between  the  circumscribed  and  inscribed  circles. 


MENSURATION. 


227 


Prob.  XXXV.  To  find  the  area  of  an  elliptic  segment,  hav- 
ing- given  the  base  of  the  segment,  its  height,  and  either  di- 
ameter of  the  ellipse,  the  base  being  parallel  to  either  diam- 
eter. 

Formulae.  —  (a)A=J  ydx,  or  -J  xdy—-  Cx(a*  — 


—  'in  (ij  '-2  (&)  ' 


(b)      A=    ydx==x(az--  **)i+a2  sin'1 


The  former  formula  of  (a)  gives  the  area  of  a  segment  whose  base 
is  parallel  to  the  conjugate  diameter  and  the  latter  the  area  of  a 
segment  whose  base  is  parallel  to  the  transverse  diameter. 

Rule.  —  Find  the  area  of  the  corresponding  segment  of  the 
circle  described  upon  the  same  axis  to  'which  the  base  of  the  seg- 
ment ts  perpendicular.  Then  this  axis  is  to  the  other  axis  as  the 
area  of  the  circular  segment  is  to  the  area  of  the  elleptic  segment. 

2,  PARABOLA. 

y  v—Zpx  is  the  rectangular  equation  of  the  parabola. 

p 

p==  -  —  -  7,  is  the  polar  equation  of  the  parabola. 
1  —  cos  6 

In  the  rectangular  equation,  HG—y,  the  ordinate;  AG=xt  the 
abscissa;  A.F=.AE-=\p.  If  any  two  of  these  are  given  the  re- 
maining one  may  be  found  from  the  equation,  p  is  a  constant 
quantity. 

Prob.  XXXVI.  To  find  the  length  of  any  arc  of  a  parabola 
cut  off'  by  a  double  ordinate. 


Formula.—  s=2j<Sdy*+dx*==       C(p  2  -f 


228 


FIXKEL'S  SOLUTION  BOOK. 


Rule.  —  When  the  abscissa  is  less  than  half  the  ordinate:  To 
the  square  of  the  ordinate  add  £  of  the  square  of  the  abscissa  and 
twice  the  square  root  of  the  sum  will  be  the  length  of  the  arc. 

I.     What  is  the  length  of  the  arc  KAff,  if  A  G  is  2  and  GH  6  ? 


By  formula,  j?== 


flag 


Since  j>/2=2/*,  we  have 


length  of  the  arc,  nearly. 
"1.  2=A  G=the  abscissa. 
2.  6=  GH=\hQ  ordinate. 


II. 


3.  36=the  square  of  the  ordinate. 


4.  y— f  of  22=4  of  the  square  of  the  abscissa. 

5.  2\f(y+36)==12.858==the  length  of  the  arc,  nearly. 
III.  .'.   12.858=length  of  the  arc,  nearly. 

Prob.  XXXyil.    To  find  the  area  of  a  parabola,  the  base 
and  height  being-  given. 

Formula.— A=2 

i.  e.,  the  area  of  parabola  HKA  is  f  of  the  circumscribed  rectangle. 

Rule. — Multiply   the  base  by  the  height  and  f   of  the  product 
will  be  the  area. 

I.  What  is  the  area  of  a  parabola  whose  double  ordinate  is  24m. 
and  altitude  16m.  ? 

By  formula,  A=%  (x.  2/)=f(  16X24 )=256m2. 

"1.  24m.=^AT"(in  last  figure )=the  double  ordinate,  or  base 
of  the  parabola. 


II. 


2.  16m.==^4  £=the  altitude  of  the  parabola. 


3.  /.  384m2==16  X  24=the  area  of  the  rectangle  circumscribed 

about  the  parabola. 

4.  f  of384m2=256m2=the  area  of  the  parabola. 
III.  .•.  The  area  of  the  parabola  is  256m2. 


MENSURATION.  229 

Prob.  XXXVIII.  To  find  the  area  of  a  parabolic  frustum 
having-  given  the  double  ordinates  of  its  ends  and  the  dis 
tance  between  them. 

B*  _  £3 

Formula.—  A=%aXj22__£2  ,  in  which  ais  the  distance  be 

tween  the  double  ordinates,  B  the  greater  and  b  the  lesser  doubl< 
ordinate. 

Rule.  —  Divide  the  difference  of  the  cubes  of  the  two  ends  by  the 
difference  of  their  squares  and  multiply  the  quotient  by  \  of  the 
attitude. 

I.  What  is  the  area  of  a  parabolic  frustum  whose  greater  base 
is  10  feet,  lesser  base  6  feet,  and  the  altitude  4  feet? 

By  formula,  A= 

(1.  10  ft.=the  greater  base, 

2.  6  ft.=the  lesser  base,  and 

3.  4  ft.=the  altitude. 

4.  784  cu.  ft.=103—  63=the  difference    of  the  cubes  of  the 
j  i      two  bases. 

'    5.  64  sq.  ft.=102  —  6  2=the  difference  of  the  squares   of  the 

two  bases. 
6.  12£  ft.=784-i-64=the  quotient  of  the  difference    of  the 

cubes  by  the  difference  of  the  squares. 
17.  .',  iX(4Xl2i)=32t  sq.  ft.=the  area  of  the  frustum. 
III.  .-.  The  area  of  the  frustum  is  32f  sq.  ft. 

3.  HYPERBOLA. 

1.  a2y2  —  62x2=  —  a-b'1  is  the  equation  of  the  hyperbola  referred 
to  its  axes  in  terms  of  its  semi-axes. 

32 

2.  y*=  --  ~(2ax  —  x2)  is  the  equation  of  a  hyperbola  referred 

to  its  transverse  axis  and  a  tangent  at  the  left  hand  vertex. 

3.  0=—        —4  is  the  polar  equation  of  the  hyperbola. 

1  —  e  cos  u 

Having  given  any  three  of  the  four  quantites  «,  b,  x,y,  the  oth- 
er may  be  found  by  solving  the  rectangular  equation  with  refer- 
ence to  the  required  quantity. 

Prob.  XXXIX.  To  find  the  length  of  any  arc  of  an  hyper- 
bola, beginning1  at  the  vertex. 


|/<Ca2_l_^2\r2    I    ^4^ 


.  _ 

--*^ 


1     <*2*2        1.1-3     ^4+4^2^2  4        1.1.3.3.5 
+1.2.3  ~bT~'    1.2.3.4.5         38       y  ^12.3.4.5.6-7 


230  FINKEL'S  SOLUTION  BOOK. 

Rule. —  !•     Find  the  parameter  by  dividing  the  square  of  the 
conjugate  diameter  by  the  transverse  diameter. 

2.  To  19  times  the  transverse,  add  21   times  the  parameter  of 
the  axis,  and  multiply  the  sum  by  the  quotient  of  the  abscissa  di- 
vided by  the  transverse. 

3.  To  9  times  the  transverse,  add  21  times  the  parameter ,  and 
multiply  the  sum  by  the  quotient  of  the  abscissa  divided  by  the 
transverse. 

4.  To  each  of  the  products  thus  found,  add  15  times  the  par a- 
qneter,  and  divide  the  former  by  the   latter;  then   this  quotient 
being  multiplied  by  the  ordinate  iv ill  give  the  length,  nearly. 

{Bonnycastle1  s  Rule*} 

NOTE. — A  parameter  is  a  double  ordinate  passing  through  the  focus. 
I.     In  the  the  hyperbola  DA  C,  the  transverse  diameter  GA 
=80,   the    conjugate    777=60,   the   ordinate   7?C=10,  and  the 
abscissa  yl^?=2.1637  ;  what  is  the  length  of  the  arc  DA  C? 

By  formula, 


1.1.3     u-    ,  ^  ~ 
1.2.3.4.5         b* 
1.13.3.5 


1.2.3.4.5.6-7 


20.658.  __^_ 

"  FIG.  20, 

1.  45=207^0^=23^  -r-tf=the  parameter  LK  which,  in 

the  figure,  should  be  drawn  to   the  right  of  DC,  to  be 
consistent  with  the  nature  of  the  problem. 

2.  1520=19X80=19  times  the  transverse  diameter. 

3.  945=21X45=21  times  the  parameter. 

4.  2465=sum  of  these  two  products. 

5.  .02704=2. 1637-f-80=quotient  of  abscissa  and  transverse 

diameter. 

6.  2465  X  .02704=66.6536=sum    of  the  products  multiplied 

by  the  said  quotient.     Also, 


7. 

45.0216.     Whence 

8.  (15x45+66.6536  )-$-(15x45+45.0216)=741.6536-r- 

720.0216=1.03004. 

9.  .-.  1.03004XlO=10.3004=length  of  the  arc  A  C,  nearly. 
III.  .-.  The  length  of  the  arc  is  10.3004. 


MENSURATION.  231 

Prob.  XIu    To  find  the  area  of  an  hyper  bola,the  transverse 
and  conjugate  axes  and  abscissa  being:  given. 

Formulae.—  (a)A=2Jydx  =&-  Cx/  (x*—  a2  )*  dx=.-x' 


;  or,  (b)  A=4xy    * 


_f  __*!_]»    J 

.5.7.  [a*+x*\         5.7. 


Rule.  —  1.  To  the  product  of  the  transverse  diameter  and 
abscissa,  add  \  of  the  square  of  the  abscissa,  and  multiply  the- 
square  root  of  the  sum  by  21. 

2.  Add  4  times  the  square  root  of  the  product  of  the  transverse 
diameter  and  abscissa^  to  the  product  last  found  and   divide   the 
sum  by  If  5. 

3.  Divide  Jf.  times  the  product  of  the   conjugate  diameter   and 
abscissa  by  the  transverse  diameter,  and  this  last  quotent   multi- 
plied by  the  former  will  give  the  area  required,  nearly.  —  Bonny- 
castlds  Rule. 

I.  If,  in  the  hyperbola  DA  C,  the  transverse  axis  A  G  is  30,  the 
conjugate  diameter  HI,  18  and  the  abscissa  AB,  10  ;  what  is  the 
area  of  the  hyperbola  DA  C  ? 


By  formula  (a),A=x'y'—ab  log*"  —~l==25y/— 


15X9  lo,e=3o0-1351o  =300 


—135  loge  3=300— 135  Xl.09861228=151.687343,/  being  found 
from  the  equation  a2/2 — &2x/<2= — a*b£ ,  in  which  «=15,  3=9 
and  *'=15+10=25. 

1.  1.    2lV30XlO+fXl02=2lV300+50(M-7=2lV371.42857 

=21X19.272=404.712,  by  the  first  part  of  the  rule. 

2.  2.    (4V30xlO+404.712)-i-75=(4xl7.3205+404.712)~ 
H  <;          75=6.3199,  by  the  second  part  of  the  rule. 

3.  3.    .'.  18X91n°X4  X  6.3199=151.6776,  by  the   third   part 

oU 
of  the  rule,  =the  area  of  the  hyperbola,  nearly. 

III.  .-.  151.6776=the  area  oftbs  •—  "T-.-"^ 


-232        .  FINKEL'S  SOLUTION  BOOK. 

Prob.  XLiI,    To  find  the  area  of  a  zone  of  an  hyperbola. 

b  r** 


Formula.— A=2-      ?***-** 

V 


'i 

in  which  (x2,y.2  ),  and  (#j, yl  )  are  the  co-ordinates    of  the  points 
<C  and  L,  respectively. 

I.  What  is  the  area  of  a  zone  of  an  hyperbola  whose  transverse 
•diameter  is  2«=10  feet,  conjugate  diameter  2£=6  feet,  the  lesser 
'double  ordinate  of  the  zone  being  8  feet  and  the  greater  12  feet? 

By  formula,  A=  •*2JK2 — x\y\ — a^°& 


But  from  the  equation,    when  jy==y2=6,    x==x2=W\/Q  and  when 
y=y1=4:,x=x1=13%.     Substituting  these  values  of  x2  andjy2» 


we  have    ^=506- 


=     50V6—  66  1—  301og<?[f(V6  +  l)]     sq.  ft. 

Prob.  XLII.    To  find  the  area  of  a  sector  of  an  hyperbola, 
K.ALO. 

Formula.—  A=at>  lo 

Rule.  —  Find  the  area  of  the  segment  A  KL  by  Prob.  XLt>  and 
subtract  it  from  the  area  oj  the  triangle  KOL. 

I.   What  is  the  area  of  the  sector  OAL  (Fig.  20)  if  OA=a=5, 
Ol=b  =3,  and  L  F=—4:  ? 


By  formula,  A=%ab  lo 

f  20)]       But  when  y==4,  ^=134.     Hence, 


MENSURATION. 

IX.     HIGHER  PLANE  CURVES. 


233- 


1.  Higher  Plane  Curves  are  loci  whose  equations  are 
above  the  second  degree,  or  which  involve  transcendental  func- 
tions, /.  £.,  a  function  whose  degree  is  infinite. 

I.     THE  CISSOID  OF  DIOCLES. 

1.  The  Cissoid  of  Diodes   is  the  curve  generated  by  the 
vertex  of  a  parabola  rolling  on  an  equal  parabola. 

2.  If  pairs  of  equal  ordinates  be   drawn  to   the  diameter  of  a. 
circle,  and  through  one  extremity  of  this  diameter  and  the  point 
in  the  circumference  through  which  one   of  the   ordinates  is  let 
fall,  a  line  be  drawn,  the  locus  of  the  intersection  of  this  line  and 
the    equal    ordinate,    or    that    ordinate    produced    is  the  Cissoid 
of  Diodes. 

x8 

3.  y2=- is  the  equation  of  the  cissoid  referred  to  rectan- 

2a — x 

gular  axes. 

p=2a  sin#  tan#  is  the  polar  equation  of  the  curve. 

Prob.  XLIII.    To  find  the  length  of  an  arc  GAP  of  the 
cissoid. 


Formula. — s=  OAP= 


r          2(3+2)         -I) 
LVs  V20—  *V80—  3*  J  ' 


I.  What  is  the  length  of  the  arc 
OAN,  in  which  case 


By    formula,    s=a  J 


Fig.  21. 


234  FINKEL'S  SOLUTION  BOOK. 

Prpb.  XLIV.  To  find  the  area  included  between  the  curve 
and  its  asymptote,  BM. 


Formula 


_ 

/**«  /*2»  i  #3        p 

.  —  A=2  I      ydx=2  I      -J-  -  dx=\  —  J 

Jo  Jo        ^&a  —  x  L. 


-        =3?r«2,  *.*.,  3  times  the  are 


of  the  circle,  OEB. 

Note.—  The  name  Cissoid  is  from  the  Greek  xtffffoeidlq,  like  ivy,  fron 
x  t  ff  a  6  c,  ivy,  eldo<;  form.  The  curve  was  invented  by  the  Greek  geometeJ 
Diodes,  A  D.  500,  for  the  purpose  of  solving  two  celebrated  problems  ol 
the  higher  geometry;  viz.,  to  trisect  a  plane  angle,  and  to  construct  twc 
geometrical  means  between  two  given  straight  lines.  The  construction  ol 
two  geometrical  means  between  two  given  straight  lines  is  effected  by  the 
cissoid.  Thus  in  the  figure  of  the  cissoid,  ED  and  OG  are  the  two 
geometrical  means  between  the  straight  lines  OD  and  JPG:  that  is, 
OD:ED\'.OG\PG.  The  trisection  of  a  plane  angle  is  effected  by  the  con- 
choid. The  duplication  of  the  cube,  i.  <?.,  to  find  the  edge  of  a  cube  whose 
volume  shall  be  twice  that  of  a  cube  whose  volume  is  given,  may  be  effected 
by  the  cissoid.  Thus,  on  KC  lay  off  CH=2BC,  and  draw  BH.  Let  fall 
from  the  point  F,  where  B  H  cuts  the  curve,  the  perpendicular  FR.  Then 
RF=&BR.  Now  a  cube  described  on  RFis  twice  one  described  on  OR', 

OR*      OR* 
for,  since  FR=y,OR=x,  and  l?R=2a—x,  we  have  RF*  —  ^^—    ^D  ,    or 

J3K        3-F-K 

\RF*=OR*.     .'.  FR*=20R*  O^  E.  D. 

2.     THE  CONCHOID  OF  NICOMEDES. 

1.  The  Conchoid  is  the  locus  formed  by   measuring,  on  a 
line  which  revolves  about   a  fixed   point   witbout  a  given  fixed 
line,  a  constant  length  in  either  direction  from  the  point  where  it 
intersects  the  given  fixed  line. 

2.  xzy2=(&-}-y)2(a2—  y8)  is  the    equation  of  the  conchoid  re- 
ferred to  rectangular  axes. 


FIG.  22. 

3.     p=b  sec  8±a\s  the  polar  equation  referred  to  polar  co-ordi- 
nates.    In  this  equation,  0  is  the  angle  PO  makes  with  A  O. 

Prob.  XLV.    To  find  the  length  of  an  arc  of  the  conchoid. 


Formula.—  s=        l+  j        <#=jVl+tan«  6  sec«<#. 


MENSURATION. 


235 


Prob.  XLVI.    To  find  the  whole  area  of  the  ciiichoid  be- 
tween two  radiants  each  making  an  angle  8  with  OA. 

Formula.—  A=2     %r*dO=b*  J*(sec  8±a)*d6= 

tan  0-{-2a2  8,  according  as  a  is  or 


62  tan 

is  not  greater  than  b. 


or 


1.  The  area  above  the  directrix  m  m'  and 
the  same  radiants  =203  log  tan  (  -r  -|-  -  j-|-<z2#. 

The  area  of  the  loop  which  exists  when  a  is 


2. 


.         o  z  i 
cos-  --2^  log 


b  is  a2 


NOTE.  —  The  name  conchoid  is  from  the  Greek,  xor%oet8£st  from 
jloyXr),  shell,  and  sT.8oq}  form,  and  signifies  shell-form.  It  was  invented 
by  the  Greek  geometer  Nicomedes,  about  A.  D.  100  for  the  purpose  of  tri- 
secting any  plane  angle.  The  trisection  of  an  angle  may  be  accomplished 
by  this  curve  as  follows:  Let  A  OH  be  any  angle  to  be  trisected.  From  any 
point,  G,  in  one  side  let  fall  a  perpendicular,  GF,  upon  the  other.  Take 
AF=2  GO,  and  with  O  as  the  fixed  point,  m  m'  as  the  fixed  line  and  /'Pas  the 
revolving  line  of  which  Piy=.a  is  constant,  construct  the  arc  of  the  con- 
choid, PAH.  Erect  BG  perpendicular  to  mm'  and  draw  BO.  Then  is 
BOA  one  third  of  HO  A.  For  bisect  B  Kat  /,  and  draw  GI.  Also  draw  IL 
parallel  to  GK.  Since  BI—IK,  BL=LG  and  GI=BI—IK^=GO.  By 
reason  of  the  isosceles  triangle  BIG,  we  have  the  angle  GIO=!2  £GBO~=. 
2LBQA.  But  l_GIO=LIQG.  •:  2£IOA^/_2QG,  or  IOA—  %/_HOA. 

C^,  E.  F. 

3.     THE  OVAL  OF  CASSINI  OR  CASSINIAN. 

1.     The  Oval  of  Cassini  is  the  locus  of  the  vertex  of  the 
triangle  whose  base  is  2a  and  the  product  of  the  other  sides  =m*. 

2. 


=m4  is  the  rectangular  equation   of  the    curve,    in    which 


3.     r4 — 2a2r2  co$26-\-a2 — m*=Q  is  the  polar  equation  of  the 
curve. 

Discussion. — If  a  be  ^>  m^ 
there  are  two  ovals,  as  shown 
in  the  figure.  In  that  case,  the 
last  equation  shows  that  if 
OPP'  meets  the  curve  in  P 
and  P',  we  have  OP.OP'= 
V<z4 — m^\  and  therefore  the 
curve  is  its  own  inverse  with 
respect  to  a  circle  of  radius= 
VV_a»4  fflHUBSM 

FIG.  23. 


FINKEL'S  SOLUTION  BOOK. 


4.     LEMNISCATE  OF  BERNOUILLI. 

1.  This  curve  is  what  a  cassinian  becomes   when  m—a. 
above    equations    then     reduce    to 

2.  ( 

3.  r 

Prob.  XLVII.  To  find  the 
length  of  the  arc  of  the  Lemnis- 
cate. 


The- 


Formula. — s== 


FIG.  24. 


C  nT^YT^V/gL-/^2^      ia     a*dr 

=  I  -J^2H  —  si  1  -  4  }dd=  I  —dd=     I   --  —  —  =-.= 
J  ^       '  r^V       a  >/         J    r  J0       Vtf4—  r4) 


*f  So  4+toAjn  +&c-}  •  When  ^=fl« 

_|_^.|.|..Tir-)-&c"|=  arc  BPA.  .'.  The  entire  length  of  the  curve  i 


Prob.  XL.  VIII.    To  find  the  area  of  the  lemniscate. 
Formula.—  A=4  hr*ct0=4a*  A7rcos26l  d0 


5.     THE  VERSIERA  OR  WITCH  OF  AGNESI. 


1.  The  Versiera  is  the   locus  of  the  extremity  of  an  ordi- 
nate  to  a  circle,   produced  until  the  produced   ordinate   is  to  the 
ordinate  itself,  as  the  diameter  of  a  circle  is  to  one  of  the  seg- 
ments into  which  the  ordinate   divides  the    diameter,  these  seg- 
ments being  all  taken  on  the  same  side. 

2.  Let  P  be  any  point  of  the  curve,  PD=y,  the  ordinate  of 
the    point    P    and 

OD=X)  the  abscis- 
sa. Then,  by  defini- 
tion, EP  \EF\\ 
AO\EO,    or  x: 
EF  :  :2a  \y.       But 


2a  :y.     Whence  x*y=  FIG.  25. 

—  y)  is  the  equation  referred  to  rectangular  co-ordinates. 


MENSURATION.  237 

3.     r(r*  —  r*  sin2  6-\-4a2)  sin6=8a*  is  the  polar  equation  of 
the  curve. 

Prob.  XLIX.    To  find  the  length  of  an  arc  of  the  Versiera. 

^ 

This  can  be  integrated   by  series   and   the   result  obtained  ap- 
proximately. 

Prob.  L.    To  find  the    area    between    the  cttrve   and  its 
asymptote. 


Formula. 


.—  A=2  A/^=2X8«3  /***  ,  f  *   a  = 
J*  Jo    x*--4a2 


20  J0 
Rule.  —  Multiply  the  area  of  the  given  circle  by  4. 

NOTE.  —  This  curve  was  invented  by  an  Italian  lady,  Dona  Maria  Agnesi, 
1748. 

6.     THE  LIMACON. 

1.  The  Limacon  is  the  locus  of  a  point  P  on  the  radius 

vector  OP)  of  a  circle  OFB  from  a  fixed 
point,  <9,on  the  circle  and  at  a  constant 
distance  from  either  side  of  the  circle. 

2.  (x*+y*—ax)*  =  l>*(x*+y*)   is 
the  rectangular  equation   of  the  curve. 

3.  r=acos9±b\*  the    polar   equa- 
tion.    In  these  equations,  a=  OA    and 


Prob.  LI.    To    find  the  length  of 
an  arc  of  the  Limacon. 

Formula. — s= 


FIG.   26. 
ibcos  fy)c 


J\  j  (a+&)2  cos^e-fA-(d—  bysin*^  \  d8.     .-.  The    rectification 


of  the  Limacon  depends  on  that  of  an  ellipse  whose  semi-axes  are 
(a-\-b)  and  (a—  b.) 

When  a=t>,  the  curve  is  the  cardioid,  the  polar  equation    of 


which  is  r=a(l+cos  6  ),  and 


=J  ^r2+(jj$ 


238 


FINKEL'S  SOLUTION  BOOK. 


OdO— 


C^Tt 

2a  I        cos  4-  6  d&=&a=the  entire  length  of  the  cardioid. 

J    7t 

Prob.  LJI.    To  find  the  areaof  the  Liimacon. 

* 


Formula.—  A= 


=\        (a  cos  0+&)*d0= 


.  When  a=b,  the  curve  becomes  a  cardioid,  and  A= 
%7ra2.  When  cT>b,  the  curve  has  two  loops  and  is  that  in  the  figure. 
r=acosO-\-b  is  the  polar  equation  of  the  outer  loop,  and  r=a 
COs  0  —  b  is  the  polar  equation  of  the  inner  loop.  The  area  of  the 

_ift 
inner  loop  is  A=J%r*d(t=%  C™      <l(acosQ—byde= 


NOTE.  —  This  curve  was   invented   bj    Blaise   Pascal  in  1643.     When  «— 
2£,  the  curve  is  called  the  Trisectrix. 

7.  THE   QUADRATRIX. 

1.  The  Quadratrix  is  the  locus   of  the  intersection,  P,of 
the  radius,     OD,   and  the  ordinate 

QN,  when  these  move  uniformally, 
so  that  ON:OA:\l_BOD:%n. 

fa  —  x  n\ 

2.  y=xtanl  -  .-   1      is     the 

rectangular  equation    of  the  curve, 
in   which    a—OA,     x=ON,    and 


3.     The  curve  effects  the    quad- 
rature of  the  circle,  for    OCiOBr. 
rc  ADB. 


Prob.  LIII.    To  find  the  area 
enclosed  above  the  x— axis. 

Formula. — A=Jydx=  FIQ  2J 

/v  tan  I  .-    }dx=4:a27r~1  log  2. 
\^    a     %  J 
•**    *      **  ^ 

NOTE. — This  curve  was  invented  by  Dinostratus,  in  370  B.  C. 

8.  THE  CATENARY. 

1.  The  Catenary'^  the  line  which  a  perfectly  flexible  chain 
assumes  when  its  ends  are  fastened  at  two  points  as  B  and  C  in 
the  figure. 


MENSURATION.  239 

a>  X 

2.  y=^a(ea-\-e~a)  is  the  rectangular  equation  of  the  curve,  in 
which a=  OA.  A  is  the  origin  of  co-ordinates.  BAC  is  the 
catenary.  M*APM\&  the 
evolute  of  the  catenary 
and  is  called  the  Tr»c- 
trix.  To  find  the  equa- 
tion of  the  curve,  let  A  be 
the  origin  of  co-ordinates. 
Let  s  denote  the  length 
of  any  arc  AE  ;  then,  if  p 
be  the  weight  of  a  unit 
of  length  of  the  chain, 
the  verticle  tension  at  E, 

is  sp.     Let  the  horizontal  FIG.  28. 

tension  at  E,  be  ap,  the  weight  of  a  units  of  length  of  the  chain. 
Let  EG  be  a  tangent  at  E,  then,  if  EG  represents  the  tension  of 
the  chain  at  B,  EF  and  GF  will  represent  respectively  its  hori- 
zontal and  its  vertical  tension  at  B. 


___        . 
'   dx~EF~ap~a     ''  a 


/•      ds 

•  •'•  x=aJ  v* 


(s-\-*  a2  +s2  )-\-c.      Since   x=o,    when   s=o,  c= 


x=a  log-+>l4--          From  this  equation,  we  find  s= 


af   to      -x*\ 

2\f"~^e  »J  which  is  the  length  of  the  curve  measured  from  A. 

dy      s         dy      s         f  *        »X 
But-/=-.  /.  -f  =-  =  41   Ji—e-a  1. 

dx      a         dx      a      ^\e  J 


Prob.  I<IV.    To  find  the  area  of  the  Catenary. 

C     ae  an 

Formula.  —  A=Jydx=J%a  \  ^+e^ 

(»  25X  :- 

ea-\-i~a  J=a\y2  —  a*.  This  is  the  area  included  between  the 

axis  of  #,  the  curve  and  the  two 


NOTE.  —  The  form  of  equilibrium  of  a  flexible  chain  was  first  investigated 
by  Galileo,  who  pronounced  the  curve  to  be  a  parabola.  His  error  was  de- 
tected experimentally,  in  1669,  by  Joachim  Jungius,  a  German  geometer; 
but  the  true  form  of  the  Catenary  was  obtained  by  James  Bernouilli,  in  1691. 


240  FINKEL'S  SOLUTION  BOOK. 

9.  THE  TRACTRIX. 

1.  The  Tractrix  is  the  involute  of  the  Catenary. 

(          . )  . 

2.  x=a  log     0-j-V(08_- y2)     —a  logy— V(«2— y2,  is  the  rect- 
angular equation  of  the  curve. 

I*rob.  LV.    To  find  the  length  of  an  arc  of  the  tractrix. 

Formula. — s—  a  log  (  -  Y 

Prob.  I/VT.    To  find  the  area  included  by  the  four  branches* 

/•« 

Formula. — A=Jydx=  — 4Jo  \a 2  — y 2  dy=7ta 2 . 

1C.  THE  SYNTRACTRIX. 

1.  The  Syntractrix  is  the  locus  of  a  point,  Q,  on  the  tan- 
gent, PT,  of  the  Tractrix. 

2.  x=  a  log  |  c+</(cz— y2)  I —a  log/— V(c2— j/2)   istherect- 

angular  equation  of   the    Syntractrix,    in    which  c  is  QT1,  a  con- 
stant length. 

11.     ROULETTES. 

1.     A.  Houlette  is  the  tocus    of  a    point    rigidly    connected 
with  a  curve  which  rolls  upon  a  fixed  right  line  or  curve. 

(a)  CYCLOIDS. 

1.  The  Cycloid  is  the  roulette  generated  by  a  point  in  the 
circumference  of  a  circle  which  rolls  upon  a  right  line. 

2.  A  Prolate  Cycloid  is  the  roulette  generated  by  a  point 
without  the  circumference   of  a   circle   which  rolls  upon  a  right 
line. 

3.  A  Curtate  Cycloid  is  the  roulette  generated  by  a  point 
within  the  circumference  of  a  circle  which  rolls  upon  a  right  line. 

4.  x±=versin~ly — *J%ry — -y2  is  the    rectangular   equation  of  the 
cycloid  referred  to  its  base  and   a    perpendicular  at  the  left  hand 
vertex.  To  produce  this  equation,  let  AN=x  and  PN=y,  P  be- 
ing any  point  of  the  curve.  Let  OC— 

r=the  radius  of  the  generatrix  OPL. 
Now  AN=A  O—NO.     But  by  con- 
struction A  O=arc  P  O=versin~v 
or  versin~*y   to    a    radius  r. 


YI   fy— 

Or,  we  may  have  x—a(  6 — sinO) ,  and  FIG.  29. 

y=a(l=cos  6)  in  which  6  is  the  angle,  PCO,  through  which  the 
generatrix  has  rolled. 


MENSURATION.  241 


For   x=AO—  NO.     But   AO=aLP  CO=aO,    and 
=  PC  sin/_PCF=a  sinQ.     .-.  x=aO—a  sin6=a(6—sin8)  ,   y= 
OC—CF=a—CF.      But  CF=PC  cos  LPCF^a  cos 
a  —  a  cosO=a(\  —  cosB). 

Prob.  LVII.    To  find  the  length  of  an  arc  of  the  cycloid. 


*&r  J  (2r—  y)~k  dy=  —  2V2r(2r—  y)*+c.  Reckoning  the  arc  from 
the  origin,  c—4r;  and  the  corrected  integral  is  s=  —  2(2r) 

(2r—  _y)*4-4r.  Whenj/=2r,  s=4r.  .-.  The  whole  length  of  the 
cycloid  is  8^=4Z?,  i  e.^  the  length  of  the  cycloid  is  4  times  the 
diameter  of  the  generating  circle. 


.  —  (1)  Multiply  the  corresponding  chord  of  the  genera* 
trix,  by  2.      To  find  the  length  of  the  cycloid  : 

(#)  Multiply  the  diameter  of  the  generating  circle  by  4- 

I.     Through  what  distance   will    a   rivet  in    the  tire  of  a  3-ft. 
buggy  wheel  pass  in  three  revolutions  of  the  wheel? 

By  formula,  5t=3(S>)=24xlift.=36ft. 

1.  3ft.=the  diameter  of  the  wheel.     Then 

2.  12ft.=4x3ft.=di*tance    through    which   it  moves  in  1 


II. 


revolution. 


3.  .-.  36ft.=3Xl2f't.=distance  through  which  it  moves  in 

3  revolutions. 
III.  .-.  It  will  move  through  a  distance  of  36  ft. 

Prob.  ITVIII.    To  find  the  area  of  a  cycloid. 

Formula.—  A=2  fyctx=2  f  '    y*-^—  = 
J  'o     V2ry—  y* 


Rule.  —  Multiply  the  area  of  the  generating  circle  by  3. 

I.  What  is  the  area  of  a  cycloid  generated  by  a  circle  whose 
radius  is  2ft.  ? 

By  formula,^4=37rr2=37r22==127T=37.6992  sq.  ft. 
(  1.  2ft.=the  radius  of  the  generating  circle. 

II.  <  2.  7r22:=12.5664  sq.  ft=the  area  of  the  generating  circle. 
(  3.  37r22=37.6992  sq.  ft.=the  area  of  the  cycloid, 

III.  .-.  The  area  of  the  cycloid  is  37.6992  sq.  ft. 

Prob.  L.IX.  Awheel  whose  radius  is  r  rolls  along-  a  hori- 
zontal line  with  a  velocity  v';  required  the  velocity  of  any 
point,  P,  in  its  circumference;  also  the  velocity  of  P  horizon- 
tally and  vertically. 


242  FINKEL'S  SOLUTION  BOOK. 

Since  a  point  in  the  circumference  of  a  wheel  describes,  in 
space,  a  cycloid,  let  P,  Fig.  29,  be  the  point,  referred  to  the 
axes  A  A'  and  a  perpendicular  at  A.  Let  (x,y)  be  the  coordi- 
nates of  the  point;  then  will  the  horizontal  and  vertical 
velocities  of  P  be  the  rates  of  change  of  x  and  y  respectively. 

\t 

O  being  the  point  of  contact,  A  O—r  versin~1—.     Since    the  cen- 

ter C,  is  vertically  over   0,  its  velocity  is    equal  to  the  rate  of  in- 
crease of  A  0.     In  an   element  of  time,  dt,  the  center  C  will  move 

(y^          rdy 
r  versiri~*J-    I—  /-  -  ==.       .'.    Its    velocity  v'  = 
rj     \2ry  —  y2 

the  distance  it  moves  divided  by  the  time  it  moves,  or  v'= 


dy  r         dy  dy      *lry  —  y2  ,       ., 

J       -r-dt=-p=  -  .  -£.  .'.  -£  =  —  J     J  T/=  the  velocity 
\2ry  _  -y2   dt  at  r 

vertically.  .  .  .  (1), 


From  the  equation  of  the  cycloid,  x=rversin~l-¥-  —  *J%ry  —  -j2,  we 

v 
have  dx=.-  —  J         dy.     Now  dx-^-dt=\hQ  velocity  of   the  point 

* 


horizontallv.   But  dx-^dt,  or—  =-.=-—  ==.  -~.     Substituting  the 

df     \/2ry—y2   dt 

value  of  ~,  we  have  —  =.-v'  .....  (2).        An    element     of   the 
dt  dt       r 

curve  APBA'  is  ds  and  this   is    the    distance  the    point    travels 

in  an  element  of  time,  dt.     .*.  •—  =  the    velocity    of  the  point,  P. 

dt 


But  ds=Vd*dx*=*  +          v'dt=          v'dt,     since, 


from  (1),  dy=  and,  from  (2),  fr.      .'.    By 


dividing  by  dt,  we  have  --=v—^-v'==  the     velocity     of     the 
point,  P  ......  (3).     From  (1),  (2),  and  (3),  we  have, 


MENSURATION.  243 

Hence,  when  a  point  of  the  circumference  is  in  contact  with 
the  line,  its  velocity  is  0\  when  it  is  in  the  same  horizontal  plane 
as  the  center,  its  velocity  horizontally  and  verically  is  the  same 
as  the  velocity  of  the  center,  and  when  it  is  at  the  highest  point, 
its  motion  is  entirely  horizontal,  and  its  velocity  is  twice  that  of 

ds        J2y          /y/2ry 

the  center.     Since  -=-  =NL—  i/  •=  -  -v'  '   we  have  by   proportion, 
dt      N  r  r 

-    :  v'  :  :  V^\r.    But  V=PF*FO*  =PO. 


.-.  The  velocity  of  P  is  to  that  of  C  as  the  chord  PO  is  to  the 
radius  CO;  that  is,  P  and  C  are  momentarily  moving  about  O 
with  equal  angular  velocity. 

(b)  THE  PROLATE  AND  CURTATE  CYCLOID. 

1.  x=a(U  —  m  sin#),  y=a(l  —  mcosO)    are    the  equations   in 
every  case. 

2.  The  cycloid  is  prolate  when  m  is  >1  as  AfP'STA',  Fig. 
30,  and  curtate  when  m  is  <C1,  as  PB.    These  equations  are  found 
thus:  Let  CP=ma,  and  tOCP=0.     Thenx=AN=AO—OW. 
But   AO=i\rc  subtended  by   /.OCP=a6,    and    ON=PCx*fa 
LNPC=ma*\K  LNPC(=£PCL=7t—(J)=ma  sin  (n—8)= 
ma  sin#.     .-.  x=atf—ma  sin6=a(6  —  m  sin#).     y=PN=OC-\~ 
PC   C05/.NPC  (=£PCL=7r—V)=a+macos(7r—0)=a— 
ma  cos^  =a(  1  —  m  cos  Q  ).     The  same  reasoning  applies  when  we 
assume  the  point  to  be  P'. 

NOTE.  —  These  curves  are  also  called    Trochoids. 
Prob.  L.X.    To  find  the  length  of  a  Trochoid. 
Formula.  —  s= 


Since  x=a(6  —  m   sin#),  dx=a(l  —  mcosP)dtf]  and  since  y= 
a(\—m  cos6>  ),  dy=am  sin^ft  /.  s= 


I.     If  a  fly  is  on  the  spoke  of  a  carnage  wheel  5  feet  in  diame- 
ter, 6  inches  up  from  the  ground,  through  what  distance  will  the 


244 


FINKEL'S  SOLUTION  BOOK. 


fly  move  while  the  wheel  makes  one  revolution  on  a  level  plane? 

Let  Cbe  the  center  of 
the  wheel,  in  the  figure, 
and  P  the  position  of  the 
fly  at  any  time.  Let  O  C= 
the  radius  of  the  carriage 
wheel  =0=2$  ft,  PC= 
2ft,  and  the  angle  OCP 
=0.  Let  (x,y)  be  the  co- 
ordinates of  the  point  P. 
Let  F,  a  point  at  the  inter-  FI&  30. 

section  of  the  curve  and  AI  be  the  position  of  the  fly  when  the 
motion  of  the  wheel  commenced.  Then  since  x—a(V  —  msinff) 
and  y=a(\  —  mcosO),  we  have  dx=a(l  —  mcoB0)dut  and  dy= 


a  m  s 


N  | 


0     dO^ 


cos 


(1+ 


snce    P    ==ma= 


,     in  which  <p=\Q.       But    7^=2  ft.,    and 

/•ITT 

ft.  ,  m==  2-7-2^=4.     •'•  ^=4  X  2^  I 

Jo 


18.84  ft. 


II.     .-.  The  fly  will  move  18.84  ft. 


Prob.  !LXI.    To  find  the  area  contained  between  the  tro- 
choid  and  its  axis. 


Formula.—  A=<tx= 


1—m  cos  6)  (  1— 


m  cos 


cos 


in6cos6)^\7r=2a2(7r-\-%m27r).     When  «w=l,  the    curve  is 

So 

the  ccloid  and  the  area=37r«2  as  it  should  be. 


*  When  <pis  replaced  by  (\n-\-<p),  this  is  an  elliptic  integral  of 
the  second  kind  and  ma    be  written 


MENSURATION. 


(c)  EPITROCHOID  AND  HYPOTROCHOID. 

1.  An  Epitrochoid  is  the  roulette  formed  by  a  circle  roll- 
ing upon  the  convex  circumference  of  a  fixed  circle,  and  carrying 
a  generating  point  either  within  or  without  the  rolling  circle. 

2.  An  Hypotrochoid  is  the   roulette   formed  by   a  circle 
rolling  upon  the    concave    circumference    of  a  fixed    circle,    and 
carrying  a  generating  point  either  within  or  without   the    rolling 
circle. 

3.  *=(#-|-£)cos  9  —  ml>cos-~-0,  y=(a-\-b)  sin  6  —  mb  sin  --  6 

are  the  equations  of  the  epitrochoids. 

In  the  figure,  let  C  be  the  center  of  the  fixed  circle  and  O  the 
center  of  the  rolling  circle.  Let  FP'Q  be  a  portion  of  the  curve 
generated  by  the  point  P'  situated  within  the  rolling  circle,  and 
let  CG=x  and  P'  G=y  be  the  co-ordinates  of  the  point,  P/  '. 

Let  A  be  the  position  of  P  when  the  rolling  commences,  and 
q*=/_POC  through  which  it  rolled.  Draw  OK  perpendicular 
to  C  G  and  P'l  perpendicular  to  OK;  draw  Z>P  and  DP'.  Let 
OP'=mOP=mb  and  the  angle  A  CD=6.  Then  x=CG=CA 
.  But  Cfe=  0C  cos  #==(04-3)  cos  0  and 


\7t-(<p+«)\=-mb 
But  arc 


AD=arcPD.    .-. 
bcp.      Whence  (p=~6. 


b 

and    x=(a-\-l>)   cos 

/>_!_/> 
mb  cos 


b 
IK=OK—OL     But 


,  and  OI=OP/  sin/  OP'I=mb  sin]  n—  (<p+B)  \ 

n  6  — 


If  f0=l,  the  point  jPx  will  be  on  the  circumference  of  the  roll- 
ing circle  and  will  describe  the  curve   APN  which  is  called  the 


246  FINKEL'S  SOLUTION  BOOK. 

Epicycloid   The  equations  for  the  Epicycloid  are  #=(tf 

£cos-^t-#,  and  y==(a4-b)  sin  d  —  £sin  —  j-  6.  The  equations  for  the 
o  o 

Hypotrochoid  may  be  obtained  by  changing  the  signs  of  b  and  mb, 
in  the  equations  for  the  Epitrochoid.     .'.  x—(a  —  £)cos  B  -\-rnb  cos 

—  —  6,  andy=(a  —  £)sin  0  —  mbsin—  7—  6  are  the  equations  for  the 

Hypotrochoid.     If  m=l,  the  generating  point  is  in  the  circumfer- 
ence of  the  rolling  circle  and  the  curve  generated  will  be  a 


a 


cycloid.  .'.  x=(a  —  b)  cos#  -\-b  cos  —  y—  '/,  and  y=(a  —  £)sin  6  — 
b  sin  —  —  9  are  the-  equations  of  the  Hypocycloid  . 

Prob.  LiXII.    To  find  the  length  of  the  arc  of  an  epitro- 
choid. 


For  mula.—  s=J*ldx*-\-dy*=          |  V—(a+b)  sin 

s  Q—  m(a+b)  cos 


a+6)  f  J 


cos 


<?0=(a+6)     i(l+m*—  2mco£  d)dB.      This 


may  be  expressed  as  an  elliptic  integral,  E(k,  cp),  of  the  second 
kind,  by  substituting  (TT  -|  —  cp)  for  0,  and  then  reducing. 

2.  By  making  /»=1,  we  have  s=(a-{-l>)\/2  f'fl^L  — 
co&y  ft)  dQ,     the  length  of  the  arc  of  an  hypocycloid. 

3.  By  changing  sign  of  b,   the  above   formula  reduces  to  «y=s 
(a  —  b)  jY(14-^2+2#2cos^  (9  )dO,  which  is  the  length  of  the  arc 
of  an  hypotrochoid. 

4.  By  making  m=l,  in  the  last  formula,  we  have  s= 

(a—  <5)V2  f(l+cos  -  &)%dd,  which  is  the  length  of  the  arc  of  «  - 
hypocycloid. 

I.     A   circle   2  ft.  in  diameter  rolls  upon  the  convex  circmii' 
ference  of  a  circle  whose  diameter  is  6  feet.     What  is  the  length 


MENSURATION.  24T 

of  the  curve  described  by  a  point  4  inches  from  the  center  of  the 
rolling  circle,  the  rolling  circle  having  made  a  complete  revolu- 
tion about  the  fixed  circle  ? 

In  Fig.  31,  let  Obe  the  center  of  the  rolling  circle;  C  the  center 
of  the  fixed  circle  ;  CZ>=3  ft.=a,  the  radius  of  fixed  circle;  OD= 
lft.=£,  the  radius  of  the  rolling  circle;  OP=4  inches=ij-  of  12 
inches—  772^  the  distance  of  the  point  from  the  center;  and  P  the 
position  of  the  point  at  any  time  after  the  rolling  begins.  Let  6= 
the  angle  A  CD  and  <p=the  angle  POD  through  which  the  roll- 
ing circle  has  rolled.  Then  we  have,  as  previously  shown,  the 
equations  of  the  locus  P, 

x=(a+b)cos  6—ml>  cos(<p-f  #)=(«-[-£)  cos  6—  mb  cos 


o 
—  mb  sin<--0=tf--3sm  6  —  mb  sin 


From  these  equations,  we  can  find  dx  and  dy, 


_ 

.-.  By  formula,  s=J\dx2+(ty2=G(a+&)J       V(l+/w2— 
2»  cos  ?  0)<*  0=24  /^VU+Ci)2—  I  cos30)«*fl=8  T 

O  J  o  Jo 

6  cos  30X<9.       Let  3  0=2$.     Then  ^=8  /^(lO—  6  cos  3 

«/0 

s^)^,  =2Hf^  [l—  i-l  cos*0—  i.i-( 

COS  ^-fi  t-  f  (I)  4  COS^- 


(|)4—  &c.     =10|7TX  .773=26.9  ft,  nearly. 


Remark.  —  When  the  point  is  on  the  circumference  of  the  roll 
ing  wheel,  the  length  of  the  curve   generated  by  the  point  is 


If  we  let  the  conditions  of  the  above  problem  remain  the  same,. 


248  FINKEL'S  SOLUTION  BOOK. 

only  changing  the  generating  point  to  the  circumference,  we  have 
for  the  length  of  the  curve,  *=6V2(3+1)  /^Vfl— cos3ff)d6= 

48 /2  V(l — i  cos 2(p )</<£>,  where  <p=f  #.     Expanding  this    by    the 
Binomial  Theorem   and   integrating    each    term    separately, 
24*     l- 


I.  A  circle  whose  radius  is  1  ft.  is  rolled  on  the  concave  cir- 
cumference of  a  circle  whose  radius  is  4  ft.  What  is  the  length 
of  the  curve  generated  by  a  point  in  the  circumference  of  the 
rolling  circle,  the  rolling  circle  having  returned  to  the  point  of 
starting? 

-.(a — £)cos  8-\-b  cos  — - —  0, 


^jr  _  ^ 

y=(a  —  b)  sin  6  —  b  sin  —  —  0,  are    the  equations    of  the    curve 

which  is  a  hypocycloid.     In  these  equations  #=4  and  3=1. 
.-.  x=3  cos  0+cos30=4cos8  0,  and 
sin  0—  sin  3  0=4  sin3  0.     Whence, 

•nd.infl=« 


cos2  0+sin2  ft=-|_  .     But  cos8  0+sm2  0=1. 


=43,  which  is  the  rectangular  equation  of  the  curve. 


By  formula,5=(^»  \-dy*  )=4 


=-6x4=24  tt 


X.     PLANK  SPIRALS. 

1.  ^4.  Plane  Spiral  is  the  locus  of  a  point  revolving  about 
a  fixed  point  and  continually  receding   from    it  in  such  a  manner 
that  the  radius  vector  is  a  function  of  the  variable  angle.       Such 
a  curve  may  cut    a    right  line    in    an    infinite  number  of  points. 
This  would  render  its    rectilinear    equation  of  an  infinite  degree. 
Hence,  these  loci  are  transcendental. 

2.  The  Measuring  Circle  is  the    circle  whose  radius  is 
the  radius  vector  of  the  spiral,  at  the  end  of  one  revolution  of  the 
generating  point  in  the  positive  direction. 


MENSURATION. 


249' 


3.     A  Spire  is  the  portion    of   the    spiral  generated  by  any 
one  revolution  of  the  generating  point. 

1.     THE  SPIRAL  OF  ARCHIMEDES. 

1.  The  Spiral  of  Archimedes    is   the  locus  of  a  point 
revolving  about  and  receding  from  a  fixed  point  so  that  the  ratio* 
of  the  radius  vector  to  the  angle    through    which   it    has   moved 
from  the  polar  axis,  is  constant. 

2.  r=aO  is  the  polar  equation  of  this  curve. 

Prob.  L.XIII.  To  find 
the  length  of  the  spiral 
of  Archimedes. 

Formula.  —  s= 


which  is  the  length  of  the 
curve  measured  from  the 
origin. 


FIG-  32. 


27r-f-V[l+(27r)2]  (  is    the  length 


of  the  curve  made  by  one  revolution  of  the  generating  point. 
Prob.  LiXIV.    To  find  the  area  of  the  spiral  of  Archimedes. 

Formula.—  A=$Jr*<t0=$azJ0*<te=la2e*=^-,  the 
area  measured  from  the  origin. 
2.     THE   RECIPROCAL  OR  HYPERBOLIC  SPIRAL. 

1.  The  Reciprocal  or  Hyperbolic  Spiral  is  the  locus. 
of  a  point  revolving  around  and  receding  from  a  fixed  point  so 
that  the  inverse  ratio  of  the  radius  vector  to  the  angle  through 
which  it  has  moved  from  the  polar  axis,  is  constant. 


2. 


'=•••  is  the  polar  equation  of  the  Hyperbolic  Spiral. 


Prob.  LXV.    To  find  the  length 
of  the  Hyperbolic  Spiral. 

Formula. — s= 


FIG.  33. 


250  FINKEL'S  SOLUTION  BOOK. 

/r'(l+02)i=log  j   0+V(~l+#2)  I  —  tf-H/1+02,    is  the  length   of 
the    spiral    measured  from  the  origin. 

Prob.  LiXVI.    To  find  the  area  of  the  Hyperbolic  Spiral. 

Formula.—  A=%Jr*  dfi=%a2J--=—^,     the     area 

measured  from  the   origin.       This    result    must  be  made  positive 
since  the  radius  vector  revolves  in  the  negative  direction. 

3.  THE  LITUUS. 

1.  The  LitUUS  is  the  locus  of  a    point   revolving   around 
and  receding  from  a  fixed  point  so  that  the  inverse   ratio  of  the 
radius  vector  to  the  square  root  of  the  angle  through  which  it  has 
anoved,  is  constant. 

a 

2.  r=\7    is  tne  equation  of  the 


Lituus. 

Prob.  LXVII.    To  find  the  length 
of  the  Lituus. 

FIG.  34 

Formula.—  s=  * 


Prob.  LXVIII.    To  find  the  area  of  the  Lituus. 

Formula.~A=^Jr^0=^C~=^  log  6. 

4.     THE  LOGARITHMIC  SPIRAL. 

1.  The  Logarithmic  Spiral  is  the  locus  genera  ted  by  a 
point  revolving  around  and  receding  from  a  fixed    point  in    such 
a  manner  that  the  radius  vector  increases  in  a  geometrical  ratio, 
while  the  variable  angle  increases  in  an  arithmetrical  ratio. 

2.  r=ae  is  the  polar  equation  of  the  Logarithmic  Spiral.     If 
is  the  base  of  a   system   of  logarithms,    this  equation  becomes 

r. 


Prob.  LXIX.    To  find  the    length     of   the    Logarithmic 
Spiral.  . 


MENSURATION.  251 

r,  where  m  is  the   modulas  of  the  system 
of  logarithms. 

Prob.  LXX.     To    find    the    area  of  the 
Logarithmic  Spiral. 

Formula. — A=$Jr*d8=-~  I  rdr= 

\mr*.  Since  m=l,  in  the  Naparian  System 
of  Logarithms,  A=%r2,  i.  e.,  the  area  is  £  of 
the  square  of  the  radius  vector. 

FIG,  35 

XI.     MENSURATION  OF  SOLIDS. 

Prob.  LXXI.    To  find  the  solidity  of  a  cube,  the  length  oi 
its  edge  being  given. 

Formula. —  F=(edge)  X  (edge )X( edge )=( edge)8. 

Rule. — Multiply  the   edge   of  the   cube   by    itself,    and  thai 
product  again  by  the  edge. 

I.     What  is  the  volume  of  a  cube  whose  edge  is  5  feet? 
By  formula,    F=(edge)3=(5)3=125  cu.  ft. 

jj   (  1.  5  ft.=the  edge  of  the  cube. 

I  2.  5X5X5=125  cu.  ft.=the  volume  of  the  cube. 

III.     .-.  The  volume  of  the  cube  is  125  cu.  ft. 

Remark. — Some  teachers  of  mathematics  prefer  to  express  th<s 
volume  by  saying  5x5X5x1  cu.  ft.  =125X1  cu.  ft=125  cu.  ft. 


Prob.  ixll.    To  find  the  volume  of  a  cube,  having  given 
its   diagonal. 

( 

Rule. — Divide  the  diagonal  by  the  square  root   of  3,    and  the 
cube  of  the  quotient  will  be  the  volume  of  the  cube. 

What  is  the  volume  of  a  cube  whose  diagonal  is  51.9615  inches? 


FINKEL'S  SOLUTION  BOOK. 


r         i      v    /^Y     /^l.eeiSY     /^51. 
By  formula,   V)  ={--  )  ^j 


27,000  cu.  in. 

II.  51.9615  in.=the  diagonal. 
2.  30  in.=51.9615  in.-M/3=»=51.9615  in.-M.73205=the  edge 
of  the  cube. 
^3.  .-.  30X30X30=27,000  cu.  in.=the  volume  of  the  cube. 

III.   /.  The  volume  of  the  cube  whose  diagonal   is  51.9615  in., 
is  27,000  cu.  in. 

Prob.  LXXII1.   To  find  the  volume  of  a  cube  whose  surface 
is  given. 

Formula 


.  —  F=y^-  j  . 


Rule. — Divide  the  surface  of  the  cube  by  6  and  extract  the 
squae  root  of  the  quotient.  This  will  give  the  edge  of  the  cube* 
The  cube  of  the  edge  will  be  the  volume  of  the  cube. 

I.  What  is  the  volume  of  a  cube  whose  surface  is  294  square 
feet? 


By  formula,    V=\  ^J^-  )  = 

243  cu.  in. 

1.  294  sq.  ft.=the  surface  of  the  cube 

2.  49  sq.  ft.=294  sq.  ft.-r  6=area  of  one  fcioe  of  the  cube, 
"  ]  3.  V49= 7  ft.=length  of  the  edge  of  the  cube. 

t.  .'.  7x7x7=343cu.  ft.=volume  of  cube- 
Ill.  .'.  343  cu.  ft.  is   the   volume  of  a   cube   whose  surface    is 
294  sq.  ft. 

Prob.  LXXIV.    To  find  the  solidity  of  a  parallelopipedon. 

Formula. —  F=/X^XA   where /^length,  £=breadth, 
and  /=thickness. 

Rule. — Multiply  the  length,  breadth  and  thickness  together. 

I.     What  is  the  volume  of  a  parallelopipedon    whose  length  is 
24  feet,  breadth  8  feet,  and  thickness  5  feet? 
By  formula,   V=lx&X £=24x8x5=9 60  cu.  ft. 

(1.  24  ft=the  length. 
2.  8  ft.=the  breadth,  and 
3.  5  ft.=the  thickness. 
4.  .-.  24X8X5=960  cu.  ft.=the  volume. 
III.     /.  960  cu.  ft.=the  length  of  the  parallelopipedon. 


MENSURATION.  253 

Prob.  LXXV.    To  find   the  dimentions  of  aparallelopite 
don,  having-  given  the  ratio  of  its  dimensions  and  the  volume. 


Formula.—  1=[  V- 

-,    and   t=$/[  V-i-(mXnXp)]p,  where  m,n,  and  p 
are  the  ratios  of  the  length,  breadth,    and  thickness  respectively. 

Rule.  —  Divide  the  volume  of  the  parallelopipedon  by  the  pro- 
duct of  the  ratios  of  the  dimensions,  and  extract  the  the  cube  root 
of  the  quotient.  This  gives  th^  G.  C.  D.  of  the  three  dimensions. 
Multiply  the  ratios  of  the  dimensions  by  the  G.  C.  D.  ,  and  the 
results  'will  be  the  dimensions  respetively. 

I.  What  are  the  dimensions  of  a  parallelopipedon  whose 
length,  breadth  and  thickness  are  in  the  ratios  of  5,  4  and  3; 
and  whose  volume  is  12960  cu.  ft.  ? 

By  formula,  /=yfl2960-H(5x4x3)]5=30  ft;  £=^[12960-- 
(5x4x3)]4=24ft.  ;  and  /=y[12960-^(5x4x3)]3=18  ft. 

1.  5=the  quotient  obtained  by  dividing  the  length  by  the 

G.  C.  D.  of  the  three  dimensions.  » 

2.  4=the  quotient  obtained  by  dividing  the  breadth  by  the 

G.  C.  D.  of  the  three  dimensions. 

3.  3=the  quotient  obtained  by  dividing  the  thickness  by 

the  G.  C.  D.  of  the  three  dimensions. 

4.  .-.  5xG.  C.  D.=the  length, 

5.  4XG.  C.  D.=the  breadth,  and 
II  A   6.  3XG-  C.  D.=the  thickness. 

7.  .-.  (5XG.  C.  D.)X(4XG.C.D.)X(3XG.C.D.)=60X 

(G.  C.  D.)3—  the  volume  of  the  parallelopipedon. 

8.  .-.60(G.  C.  D.)3=12960cu.  ft. 

9.  (G.  C.  D.)3=12960-7-60=216. 

10.  /.  G.  C.  D.—  f  216=6. 

11.  .'.  5X(G.  C   D.)=5X  6=30  ft.=the  length, 

12.  4X(G.  C.  D.)=4X6=24  ft.=the  breadth,  and 

13.  3X(G.  C.  D.)=3X6=18  ft.=the  thickness. 

III.  .-.  30  ft.,  24  ft.,  and  18  ft.  are  the  dimensions  of  the  par- 
allelopipedon. 

Prob.  LXXVI.    To  find  the   convex  surface  of  a  prism. 

Formula.—  S=pXa,  in  which/  is  the  perimeter  of  the 
base  and  a  the  altitude. 


Rule.— Multiply  the  perimeter  of  the  base  by  the  altitude. 


254  FINKEL'S  SOLUTION   BOOK. 

I.  What  is  the  convex   surface  of  the  prism  ABC — Z?,  if  the 
altitude  AE  is  12  feet,  AB,  6  feet,  A  C,  5  feet,  and  BC,  4  feet.? 

By  formula,  5=«X/=12X(9+5+4)=180  sq.  ft. 

(  1.   12  ft  =the  altitude  of  the  prism. 

II.  <  2.  6  ft. +5  ft.+4  ft.=15  ft.=the  perimeter  of  the  base. 

(  3.   .'.  12x15=180  sq.  ft.=the  convex  surface  of  the  prism. 

III.  .-.   The  convex  surface  of  the  pris'm  is  180  sq.  ft. 

Remark. — If  the  entire  surface  is  required;  to  the  convex  sur- 
face, add  the  area  of  the  two  bases. 

Formula. —  T=S-\-flA<  where  2A    is    the  area    of  the 
base,  S  the  convex  surface,  and  7*  the  total  surface. 

Prob.  LiXXVII.  To  find  the  volume  of  a  prism. 

Formula. —  F— #XA  where  A  is    the  area  of  the  base, 
<z,  the  altitude. 

Rule. — Multiply  the  area  of  the  base  by  the  altitude. 

I.  What  is  the  volume  of  the  triangular  prism  ABC — Z>, 
whose  length  AE  is  8  feet,  and  either  of  the  equal  sides  AB> 
BC,  or  AC,  2£  feet? 

By  formula,    F===«X^=^SX{(2i)tiV3]==i2^5=21.6606  cu.  ft- 

1.  8  ft.=the  altitude  AE. 

2.  21  t"t.=the  length  of  one   of  the  equal 

sides  of  the  base,  as  AB. 
II.<{3.   (2|)2iV8=the  area  of  the  base  ABC, 

by  Prob.  XL 

4.  .-.  8X(2J)2iV3=12\/3=21.6506cu.  ft. 
=the  volume  of  the  prism. 

III.     .'.  -21.6506    cu,  ft.=the  volume  of  the 
prism. 

FIG.  36. 

1.     THE  CYLINDER. 
Prob.  LXXVIII.    To  find  the  convex  surface  of  a  cylinder. 

Formula. — S=^a  X  C,  in  which  a  is  the  altitude  and  C  the 
circumference  of  the  base. 

Rule. — Multiply  the  circumference  of  the  base  by  the  altitude. 

I.  What  is  the  convex  surface  of  the  right  cylinder  A  GB — C, 
whose  altitude  EF  is  20  feet  and  the  diameter  of  its  base  AB  is 
4  feet? 


MENSURATION. 


255 


By  formula,  5=aXC=20x47r=80^=251. 32736  sq.  ft 

1.  20  ft.=the  altitude  EF. 

2.  4  ft.=the  diameter  AB  of  the  base. 

3.  12.566368  ft.=47T=4x3.141592=the 

circumference  of  the  base. 
4  /.    20x12.566368—251.32736   sq.  ft.= 
the  convex  surface  of  the  cylinder. 

The  convex  surface  of  the  cylinder  is 


II. 


III. 


251.32736  sq.  ft. 

Remark. — If  the  entire  surface  is  required;  to 
the  convex  surface,  add  the  area  of  the  two 


Formula.—  T^S+IA^ZnaR+lnR*.  FIG.  37. 

Prob.  LXXIX.    To  find  the  solidity  of  a  cylinder. 
Formula.  —  K—  ^ 


in   which    A   is   the   area  of  the 


base. 


Rule.  —  Multiply  the  area  of  the  base  by  the  altitude. 

I     What  is  the  solidity  of  the  cylinder  AGB  —  C,  whose  alti- 
tude FE  is  8  feet  and  diameter  AB  of  the  base  2  feet? 


By   formula,     F=aX^== 
25.132736  cu.  ft. 

II.  8  ft.=the  altitude,  EF. 
-2.  2  ft.=the  diameter,  AB,  of  the  base. 
3.  3.141592  sq.  ft.=7r/?2:=:7rl2=  area  of  the  base. 
4.  .-.  8X3.141592=25.132736  cu.  ft. 

III.     .'.  25.132736  cu.  ft.  is  the  volume  of  the  cylinder. 
2.  CYLINDRIC  UNGULAS. 

1.     A  Cylindric  Ungula  is  any  portion  of  a  cylinder  cut 
off  by  a  plane. 

Prob.  LXXX.  To  find  the  convex  surface  of  a  cylindric 
ung-ula,  when  the  cutting-  plane  is  parallel  to  the  axis  of  the 
cylinder. 


Formula.—  S=a  J^^ 

arc  of  the  base. 


sn 


Rule.  —  Multiply  the  arc  of  the  base  by  the  altitude. 

I.  What  is  the  surface  of  the  cylindric  ungula  API  —  Q, 
whose  altitude  AD  is  32  feet  and  height  A  T  of  the  arc  of  the 
base,  2  feet  and  cord  PI  of  the  base  12  feet? 


256  FINKEL'S  SOLUTION   BOOK. 

By  formula,  S=  a^arc  PAI=aZr  s\n~1^=aX^  sin"1  (^ 

J  y  \       r 


2AT 


sq.  ft,  nearly. 


The  arc  corresponding   to  the  sin  4   is  found   from   a   table   of 
natural  sines  and  cosines  to  be  (36° 


"1  77^1 
of  2  T  or  »'    '    TT. 


a.  2  ft=the  height  u4  7W  the  arc 
2.  12  ft.=  the  length  of  the    chord    PL 


3.  12.87  ft.=2V62+22x(l+ 


arc  P^7,  by  Prob.  XXV. 

4.  ..  32X12.87=411.84  sq.   ft.  ==  convex 
surface  PAI—D. 

III.  .'.  The  convex  surface   of  the   cylindric 
tmgula  PAI—Q  is  411.84  sq.  ft. 

Remark.  —  r  is  found,  by  Prob.  XX,  formula 


FIG.  38. 


Prob.  LXXXI.    To  find  the  volume  of  a  cylindric  ungrila, 
whose  cutting  plane  is  parallel  to  the  axis. 


r*—  ? 


Formula.—  V=2  C  T^2^*  Fdydxdz—  2ay( 
Jo  J  o  Jo 


a\r2  sin"1^—  y(r*—t  y2)*\  ,  in  which  j|/  is   half   the    chord    of  the 

base.     In  this  formula  f  r2  sin"1  -  j    is   the   area   of  the  sector 

APEIA,  3indy(r2—  y2  )*  is  the  area  of  the  triangle  PEI  formed 
by  joining  the  center  E  with  P  and  /. 

Rule*  —  Multiply  the  area  of  the  base  by  the  altitude. 

I.     What  is  the  volume  of    the  cylindric   ungula  PIA  —  Z>,  if 
PI  IB  12  feet,  AT2  feet,  and  altitude  AD  40  feet? 


MENSURATION.  257 

By  formula,  V=aA—a  \r*  sin  1<^— y(rz— y2)  |  =40|l02sin'll 


—  6(102-  62  )t=4COOsin  '  |—  1920=4000^—  1920= 
2574.016-  1920=654.016  cu.  ft. 

1.  40  ft.=the  altitude  AD. 

2.  2  ft.=the  height  A  T  of  the  arc  of  the  base. 

3.  12  ft.=the  chord  P/of  the  base. 

II.<J4  16|sq.  ft.=^2+l°f(2Xl2)=the   area  of  the  base, 

by  rule,  Prob.  XXVIII. 

5.  .-.  40Xl6i=653£  cu.  ft.=the  volume  of   the  cylindrical 
ungula  PIA—D. 

III.     .'.  653^  cu.  ft.=the  volume  of  the  cylindrical  ungula. 

Remark.  —  A  nearer  result  would  have  been  obtained  by  finding 
the  length  of  the  arc  PAI  and  multiplying  it  by  half  the  radius. 
This  would  give  the  area  of  the  sector  IE  PA.  From  the  area  of 
the  sector  subtract  the  area  of  the  triangle  PIE  formed  by  join- 
ing P  and  /with  E,  and  the  remainder  would  be  the  area  of  the 
segment  PI  A. 

Prob.  tiXXXH.  To  find  the  convex  surface  of  a  cylindrio 
nngula,  when  the  plane  passes  obliquely  through  the  op- 
posite sides  of  the  cylinder. 

Formula.  —  S=^(a-\-  a')2?rr,  where  a  and  a  are  the 
least  and  greatest  lengths  of  the  ungula  and  2  TIT  the  circumfer- 
ence of  the  base  of  the  cylinder. 

Rule.  —  Multiply  the  circumference  of  the  base  by  half  the 
sum  of  the  greatest  and  least  lengths  of  the  ungula. 

I.  What  is  the  convex  surface  of  the  cylindric  angula  A  KB  A 
—NM,  if  ^4^Vis8  feet,  BM  12  feet  and  the  radius  BE  of  the 
base  3  feet? 

By   formula,    S=$(a-\-a')27rr=7r(a+a')  r=7e(S+12)  X3=* 
188  49552  sq.  ft. 

1.  8  ft—  the  least  length  A  JV  of  the  ungula,  and 

2.  12  ft.=the  greatest  length  BM. 

3.  10ft.=J(8ft.+12  ft.)=half  the  sum    of  the   least   aua 

greatest  lengths. 

4.  18.849552  ft.=67r=the  circumferenc  of  the  base. 

5.  .-.  10X18.849552=188.49552  sq.  ft.=the  convex  surface, 


258  FINKEL'S  SOLUTION  BOOK. 

III.     .'.  188.49552  sq.  ft.=the  convex  surface  of  the  ungula. 

Prob.  LXXXIII.  To  find  the  volume  of  a  cylindric  ungula, 
when  the  plane  passes  obliquely  through  the  opposite  sides 
of  the  cylinder. 

Formula.—  V=a-a/7tr'1 


Multiply  the  area  of  the  base,  by    half  the    least   and 
greatest  lengths  of  the  ungula. 

I.  What  is  the  volume  of  a  cylindric  ungula  whose  least 
length  is  7  feet,  greatest  length  11  feet,  and  the  radius  of  the 
base  2  feet? 

By  formula,  K=J(«+«')7rr2=  i(7+ll)7r22=113.097312  cu.  ft. 


II.  • 


1.  7  ft.=the  least  length  of  the  ungula,  and 

2.  11  ft.=the  greatest  length. 

3.  9  ft.=J(7  ft.+ll  ft.)=half   the  length  of  the  least  and 

greatest  lengths. 

4.  12.566368  sq.  ft.=7r22=the  area  of  the  base. 

5.  .'.9X12.566368=113.097312  cu.    ft.=the   volume  ot  the 
H     ungula. 


III.     .-.  The  volume  of  the  ungula  is  113.097312  cu.  ft. 

Prob.  L.XXXIV.  To  find  the  convex  surface  of  a  cylindric 
ungula,  when  the  plane  passes  through  the  base  and  one  of 
its  sides. 


rdx 


(r  —  £)vers-1-  | 


Rule.  —  Multiply  the  sine  of  half  the  arc  of  the  base  by  the 
diameter  of  the  cylinder,  and  from  the  product  subtract  the  prod- 
uct of  the  arc  and  cosine;  this  difference  multiplied  by  the  quo- 
tient of  the  height  divided  by  the  versed  sine  'will  be  the  convex 
surface. 

I.     What  is  the  convex  surface  of  the  cylindric  ungula  A  CB  — 


MENSURATION. 


259 


D,  whose  altitude  BD  is  28  feet,   height  BM  of  arc   of  base  4 
feet  and  chord  A  C  16  feet? 


By  formula,  5=2r 


(r-i 


28 


,  =2XlOX 


j  [V2X10X4—  42—  (  10—  4)  vers-'Aj  ,  = 


367200' 
==140[8— 6.5638]=341.068  sq.  ft. 


1.  28  ft.=the  altitude  BD. 

2.  4  ft.=the  height  BM  of  the  arc  of  the  base. 

3.  16  ft.=the  chord  A  C  of  the  arc  of  the  base. 

4.  8  ft.=the  sine   CM  of  the  arc. 

5.  10ft—  (82+42)-h-(2x4)=the  radius  OC=OB  of  the 

base,  by  Prob.  XX,  formula  /?=(08+c2)-i-2a. 

6.  6  ft.=10  ft—  4  ft.=cosine  OM  of  the  arc. 

7.  160  sq.  ft.=20x8=sine   multiplied   by  the  diameter  of 

the  base. 


II. 


8.  18.5438 


l=the    arc 


CBA,  by  formula  of  Prob.  XXV. 

9.  /.  111.2628  sq.  ft.==6Xl8.5438=the   arc  multiplied    by 
the  cosine  OM. 

10.  160  sq.  ft—  111.2628   sq.  ft.=48.7372sq.ft.=the   differ- 

ence. 

11.  .-.  341.1604  sq.  ft.=(28-5-4)X  48.7372  sq.   ft=the  con- 

vex surface. 

III.     .-.  The  convex  surface  is  341.1604  sq.  ft.  nearly. 

NOTE.  —  The  difference  in  the  two  answers  is  caused  by  the   length  of  the 
arc  CBA,  in  the  solution,  only  being  a  near  approximation. 


*  Demonstration.  —  In  the  figure,  let   BK=x,  BM=b, 
and  the  angle  BMD=S.     Then  MK=b—x,  and  IK=FL=MK 

^ 


tan  6=(b—  x)  tan  8.     But  tan^==.     .'.  FL=(b—x). 


Now  if  we  take  an  elemennt  of  the  arc  LBH,  and  from    it  draw 
a  line  parallel  to  FL,   we    will  have    an    element   of     the    sur- 


260  FINKEL'S  SOLUTION  BOOK. 

face  LBHEGF.      This   will   be  a  rectangle   whose   length  is 
~(b  —  x)  and  width  an  element  of  the    arc  LBH.     An  ele- 


ment of  the  arc  is  ds=*(dx*+dyz).      Let   HK=y.      Then^2 
2rx  —  #2,  by  a  property  of  the  circle,    from    which    we  find 


*»_  y  T'fLX 

dx.     .'.  //.?==  .'.  The  area  of  the  element  of 


a 
the  surface  is  -(b  —  x  )  /_.       ^^^  and  the   whole  surface  of  AB  C— 

^          '  2 


x  a  /* 

,^=flr-i 
—  x2  Wo 


2  (  r—  &  )  r  vers-1        . 


Prob,  LXXXV.  To  find  the  volume  of  a  cylindric  ungula, 
when  the  cutting1  plane  passes  throug-h  the  base  and  one  of 
its  sides. 

/•&  a  /•&  _ 

—  V=  I    (b—x)dA=-l  I    (l>—x)2\2rx—x2(tx, 

Jo  vJo 


.  When  x=0, 
=0.          .-.    C=—  i;rr2 


Rule. — From  f  of  the  cube  of  half  the  chord  of  the  base,  sub- 
tract the  product  of  the  area  of  the  base  and  the  difference  of  the 
radius  of  trie  base  and  the  height  of  the  arc  of  the  base;  this  dif- 
ference multiplied  by  the  quotient  of  the  altitude  of  the  ungula  by 
the  height  (versed  sine}  of  the  arc  of  the  base,  'will  give  the  vol- 
ume. 

I.  What  is  the  volume  of  a  cylindric  ungula,  whose  altitude 
BD  is  8  feet,  chord  A  C  of  base  6  feet,  and  height  BMvl  arc  of 
base  1  foot? 

By  formula,   V=-  \  f(2r£— £2)f— (r— b}  [~ ^rr2  — 


MENSURATION.  261 

-r*  sin  -  =8     f  (2X5X1-1  )f— 

18— 


(5—  l)[i«-5a—  4V2X5X1—  1—  52  sin"1  ^Q  j  ,  =8  j 


12—25  sin-1!"!  I  =528+800  sin'1  ~200*= 
o  J  )  o 

13.20394  cu.  ft. 

1.  8  ft=the  altitude 

2.  1  ft.=the  altitude  BM  of  the  arc  ABC  of  the  base. 

3.  6  ft.=the  chord  A  C  of  the  base. 

4.  18  cu.  ft=  f  of  33=|  of  the  cube  of  the  sine  of  half  the 

arc  of  the  base. 
I3 

n.< " 

ula,  (J),  Prob.  XXVIII. 

6.  16£  cu.  ft.=4  X  4TV=the  area  of  the  base  X  OM,  the  cosine 

of  the  arc  CHB. 

7.  /.  8(18  cu.  ft— 16£  cu.  ft.)=13i  cu^  ft.=the  volume  of 

the  cylindric  ungula  A  CB — D. 
III.     .\  The  volume  of  the  cylindric  ungula  A  CB— D  is  134 
cu.  ft.,  nearly. 

Prob.  LXXXVI.    To  find  the  convex  surface  of  the  frustum 
of  a  cylindric  ung-ula. 


5.  4TV  sq.   ft=fr—  o+fof  6Xl=area  of  the  base,  by  form- 


Formula.—  S= 


Rule.  —  (  1  )  Conceive  the  section  to  he  continued,  till  it  meets 
the  side  of  the  cylinder  produced;  then  say,  as  the  difference  of 
the  heights  of  the  arcs  of  the  two  ends  of  the  ungula,  is  to  the 
height  of  the  arc  of  the  less  end,  so  is  the  height  of  the  cylinder  to 
the  part  of  the  side  produced. 

(2)  Find  the  surface  of  each  of  the  ungulas,  thus  formed, 
fy  Prob.  LXXXIy.,  and  their  difference  will  be  the  convex  sur- 
face of  the  frustum  of  the  cylindric  ungula. 

Prob.  LXXXVII.  To  find  the  volume  of  a  frustum  of  a 
cylindric  uiig*ula. 

Formula.—  K=i(2rJ—  32)f—  (r—  b)    1^2— 


262  FINKEL'S  SOLUTION  BOOK. 


.  —  Find  the  volume  of  the  ungula  whose  base  is  the  the 
upper  base  of  the  frustum  and  altitude  that  as  found  by  (1)  of  the 
last  rule.  Also  the  volume  of  the  ungula  whose  base  is  the  lower 
base  of  the  frustum  and  altitude  the  sum  of  the  less  zingula  and 
altitude  of  the  frustum.  Their  difference  will  be  the  volume  of 
the  frustum. 

3.  PYRAMID  AND  CONE. 

Prob.  LXXXVIII.  To  find  the  convex  surface  of  a  right 
cone. 

Formula.  —  S=  Cxi/*=27Tr  x  %Va2-{-r*  ,.  where  C  is  the 
circumference,  h  the  slant  height,  r  the  radius  of  the  base,  and  a 
the  altitude. 

Rule.  —  Multiply  the  circumference  of  \  the  base  by  the  slant 
height  and  take  half  the  product.  Or,  if  the  altitude  and  radius 
of  the  base  are  given,  multiply  the  circumference  of  the  base  ly  the 
square  root  of  the  sum  of  the  squares  of  the  radius  and  altitude, 
and  take  half  the  product. 

I.  What  is  the  convex  surface  of  a  right  cone  whose  altitude 
is  8  inches  and  the  radius  of  whose  base  is  6  inches?. 


By  formula,  S=  2^rXiVa2+r2 
188.495559  sq.  in. 

1.  6  in.=the  radius  AD  of  the  base, 

and 

2.  8  in.—  the  altitude  CD. 

3.  10  in.=Vr82+^2=the    slant 

height  CA. 
11.14.  37.6991118  in.=2frr=12x 

3.14159265=the  circumfer- 
ence of  the  base. 

4.  .-.  188.495559  sq.  in= 

i(10x37.6991118)=the   con- 

vex surface  of  the  cone.  p,Q 

III.     .'.  The  convex  surface  of  the  cone  is  188.495559  sq.  in. 

Prob.  LXXXIX.     To  find  the  convex  surface  of  a  pyramid. 

Formula.  —  S=$pXb,   in  which  p  is  the   perimeter  of 
the  base  and  h  the  slant  height. 

Rule.  —  Multiply  the  perimeter  of  the  base  by  the  slant  height 
and  take  half  the  product. 


MENSURATION.  26a 

I.     What  is  the  convex  surface  of  a  pentagonal  pyramid  whose 
slant  height  is  8  inches  and  one  side  of  the  base  3  inches? 

By  formula,  fc^/  X/*=i(  3+3+3+3+3  )  X  8=60  sq.  in. 

1.  8  in.=the  slant  height. 

2.  3  in.—  the  length  of  one  side  of  the  base. 
II..J3.  5X3  in.  =15  in.=the  perimeter  of  the  base. 

4.  .'.  1(15X8)=60  sq.  in.=the  convex  surface  of  the  pyra- 

mid. 
III.     .*.  The  convex  surface  of  the  pyramid  is  60  sq.  in. 

Remark.  —  If  the  entire  surface  of  a  pyramid  or  cone  is  required, 
to  the  convex  surface  add  the  area  of  the  base. 

Formula.  —  T=S-\-A,  where  A    is  the  area  of  the  base 
and  S  the  convex  surface. 

Prob.  XC.    To  find  the  volume  of  a  pyramid  or  a  cone. 

Formula.  —  V=^aA=^aX7tr2,  where    a  is  the  altitude 
and  A=7tr2  the  area  of  the  base. 

Rule.  —  Multiply  the  area  of  the  base  by  the  altitude  and  take 
one-third  of  the  product. 

I.     What  is  the  volume  of  a    cone    whose    altitude    CD  is  18 
inches  and  the  radius  AD  of  the  base  3  inches? 


By  formula,   V=^a  X?rr2=4x  18  X  7r32=54x  3.14159265= 
169.646  cu.  in. 

1.  18  in.=the  altitude  CD,  and 

2.  3  in.=the  radius  AD. 


II. 


3.  28.27433385  sq.  in.=z:7rr2=327r=the  area  of  the  base. 


4.  .'.  169.6460031  cu.   in.=£n^4=£xl8x387r=the   volume 

of  the  cone. 
III.     .-.  The  volume  of  the  cone  is  169.6460031  cu.  in. 

Prob.  XCI.    To  find  the  convex  surface  of  a  frustum  of  a 
cone. 

Formula.—  S= 


+  (r  —  r')2,  in  which  C  is  the  circumference  of  the 
lower  base,  Cx  the  circumference  of  the  upper  base,  and  ^,= 
Va*+(r—  r'}*  ,  the  slant  height. 

Rule.  —  Multiply  half  the  sum  of  the  circumferences  of  the 
two  bases  by  the  slant  height. 

1.  What  is  the  convex  surface  of  the  frustum  of  a  cone  whose 
altitude  is  4  feet,  radius  of  the  lower  base  4  feet,  and  the  radius  of 
the  upper  base  1  foot? 


•264  FINKEL'S  SOLUTION  BOOK. 


By  formula,  5=7i'(r+r/)Va2+(r~r/)2=^(4+l)\/42+(4— 
.=257r.=78.539816  sq.  ft. 

1.  4  ft.=  the  altitude  OE, 

2.  4  ft.=the  radius  AE  of  the  lower 

base,  and 

3.  1  ft.=the  radius  DO  of  the  upper 

base. 

4.  3  ft.=^4^—  PE(=DO)=r—  r>. 

5.  5  ft= 


=AD,  the  slant  height. 

6.  87r=the  circumference  A  GJBffot 

the  lower  base. 

7.  2w=the  circumference  DIC  of  the  upper  base. 

8.  57r=i(87r-{-27r  )=half  the  sum  of  the  circumferences. 

9.  .-.  5x57r=257T=78.539816sq.  ft.=the  convex  surface  of 

the  frustum. 
III.     .-.  The  convex  surface  of  the  frustum    is  78.539816  sq.  ft. 

Remark. — If  the  entire  surface  of  the    frustum    is   required,  to 
the  convex  surface  add  the  area   of  the  two  bases. 

Formula.—  T=S+A+A/=7r(r+r/  )*la*+(r—r')*  + 


Prob.  XGII.    To  find  the  convex  surface  of  the  frustum  of 
a  pyramid. 

Formula.—  S= 


Rule.  —  Multiply  half  the  sum  of  the  perimeters  of  the  two 
bases  ~by  the  slant  height. 

I.  What  is  the  convex  surface  of  the  frustum  of  a  pentagonal 
pyramid,  if  each  side  of  the  lower  base  is  5  feet,  each  side  of  the 
upper  base  1  foot,  and  the  altitude  of  the  frustum  10  feet? 

Before  we  can  apply  the  formula,  we  must  find  the  slant 
height.  Produce  FO,  till  OK=OE.  Divide  OK  into  extreme 
and  mean  ratio  at  H.  Draw  EH.  Then  KO  :  OH'.  :  OH  :  KH. 


whence  OH*+KOx  OH^KO*.     Completing  the  square  of  this 
equation,  OH*+KC>X  OH+lKQt^KO*  ,  from  which  OH(= 


MENSURATION. 


265 


.  But  EF= 


2V;5),  and   te 
where  5  is   a  side  of  the  lower  hase,= 


may  be  considered    the   radius    R    of   a  circum- 
VlO—  2V5* 

scribed  circle  of  the  lower  base.     In  like  manner,  the  radius  r  of 
the  circumscribed  circle  of  the    upper    base    may    be  found  to  b 

__Jfl       ,  where  s'  is  a  side  of  the  upper  base,—-—       ==. 
VlO—  2V5  V5 


the  apothem  of  the  lower  base,^ 


=|V650+lWl==the  slant  height. 


II. 


By  formula,  S=£(25+5; 

155.5795  sq.  ft 

1.   10  ft.=the  altitude  oO. 

5  fa=EA,  one  of  the  equal    sides   < 

the  lower  base. 
1  K.=ed9    one  of  the  equal    sides    of 

the  upper  base. 

4.  f  V650+W1=/^  the  slant  height. 
5X5  ft.=25  ft.=the  perimeter  of  the 
lower  base. 

6.  5X1  ft.=5  ft.=the  perimeter    of 

upper  base. 

7.  .-.    -H  25+5  )|V650+10Vl  =  155.5795  sq.ft.=the  conve, 

surface. 

III.     /.  The  convex  surface  of  the  frustum  is  155.5791 
Prob.  XCIII.    To  find  the  volume  of  a  frustum  of  apyia- 
mid  or  a  cone. 


266  FINKEL'b  SOLUTION  BOOK. 

Formula.—  (a)  V=%a(A+VAA~'+A'),  in  which  A 
is  the  area  of  the  lower  base,  A'  the  area  of  the  upper  base  and 
V  '  AA'  the  area  of  the  mean  base.  When  we  have  a  frustum  of 
a  cone,  (b)  V=$a(A+»fAA'+A')=$a(  7tR*+V(  xR*  X  nr*  )+ 


Rule.  —  (1)  Find  the  area  of  the  mean  bast  by  multiplying  the 
area  of  the  upper  and  lower  bases  together  and  extracting  the 
square  root  of  the  product. 

(2)  Add  the  upper,  lower,  and  mean  bases  together  and  multi- 
ply the  sum  by  -J  the  altitude. 

I.  What  is  the  solidity  of  a  frustum  of  a  cone  whose  altitude 
is  8  feet,  the  radius  of  the  lower  base  2  feet,  and  the  radius  of 
the  upper  base  1  foot? 

By  formula  (3),  V^Tta^+Rr+r*)^*  8(4+2+1)= 
£X  56  TT  =58.6433  cu.  ft. 

1.  8  ft=the  altitude. 

2.  2  ft.=the  radius  of  the  lower  base. 

3.  1  ft.=the  radius  of  the  upper  base. 

4.  4^=the  area  of  the  lower  base. 

5.  7r==the  area  of  the  upper  base. 

6.  27r=V47r  X  7f=the  area  of  the  mean  base. 

7.  4  7t  +  7T+2  n==l  7t  =the  sum  of  the  areas  of  the  three  bases. 
.'.  itX8x7#  =58.6433  cu.  ft.=the  solidity  of  the  frustum, 

III.     .-.  Ths  solidity  of  the  frustum  is  58.6433  cu.  ft 

4.    CONICAL  UNGULAS. 

1.  A  Conical  Unyula  (  Lat.  ungula,  a  claw,  hoof,  from 
unguis,  a  nail,  claw,  hoof)  is  a  section  or  part  of  a  cone  cut  ofl 
by  a  plane  oblique  to  the  base  and  contained  between  this  plane 
and  the  base. 

Prob.  XCIV.    To  find  the  surface  of  a  conical  imgula. 
Formula.—  S= 


II, 


i-(2/?—  t\r—  (^?+r—  t)x~\  )    , 
2*  cos"1  '      ^     '  I  >  dx,  where  a  is  the  altitude 


of  the  ungula,  R  the  radius  of  the  base,  ^the  radius  of  the  upper 
base  of  the  frustum  from  which  the  ungula  is  cut,  /  the  distance 
the  catting  plane  cuts  the  base  from  the  opposite  extremity  of 
the  base,  and  x  the  radius  of  a  section  parallel  to  the  base  and  at 
a  distance  h  —  y  from  the  base. 


MENSURATION.  267 

I*  rob.  XCV.    To  find  tlie  volume  of  a  conical  uiigula. 


CR 
Formula.  —  F=  /    A  dy= 


(R—r)x 


—  t)(R+r—  t) 


dx, 

where  the  letters  represent  the  same  value    as    in    the    preceding 
problem  and  dy=(  —  --  \dx,  since  y=  —  ^  --  -. 

^L    £\.  —  —"?*    J  _/\     '  ••  /* 

Prob.  XCVT.  To  find  the  convex  surface  of  a  conical  un- 
JMI  la,  when  the  cutting1  plane  passes  through  the  opposite 
extremities  of  the  ends  of  the  frustum. 


Formula.  —  S= 


This  formula  is  obtained  by  putting  t  —0,  in 
the  formula  of  Prob.  XCIV.,  and  integrating  the 
result.  For,  in  this  problem,  the  cutting  plane 
AHCK  passes  through  the  opposite  point  A,  and 
therefore  the  distance  from  A  to  the  cutting 
plane  is  0.  .-.  t—0. 


FIG.  43. 


Rule.  —  Multiply  half  the  sum  of  the  radii  of  the  bases  by  the 
square  root  of  their  -product  and  subtract  the  result  from  the 
square  of  the  radius  of  the  lower  base.  Multiply  this  difference 
by  7t  times  the  slant  height  and  divide  the  result  thus  obtained  by 
the  difference  of  the  radii  of  the  bases. 

Prob.  XCVII.  To  find  the  volume  of  a  conical  ungrula, 
when  the  cutting-  plane  passes  through  the  opposite  extremi- 
ties of  the  ends  of  the  frustum. 


Formula.- l^^^,  (&-**) 
3(/c — r)  V  J 


This  formula  is  obtained  by  putting  t=0,  in  the  formula  of  Frob. 
XCV.,  and  integrating  the  result. 


.  —  Multiply  the  difference  of  the  square  roots  of  the 
cubes  of  the  radii  of  the  bases  by  the  square  root  of  the  cube  of  the 
radius  of  the  lower  base  and  this  product  by  \n  times  the  altitude. 


268  FINKEL'S  SOLUTION  BOOK. 

Divide  this  last  product  by  the  difference  of  the  radii  of  the  two 
bases  and  the  quotient  'will  be  the  volume  of  the  ungula. 

I.  A  cup  in  the  form  of  a  frustum  of  a  cone  is  7  in.  in  diame- 
ter at  the  top,  4  in.  at  the  bottom,  and  6  in.  deep.  If  when  full 
of  water,  it  is  tipped  just  so  that  the  raised  edge  of  the  bottom 
is  visible;  what  is  the  volume  of  the  water  poured  out? 


By  formula,  F^7?—==|^(  49-8^7)= 
102.016989  cu.  in. 

Remark.  —  Fig.  43  inverted  represents  the  form  of  the  cup  and 
APBQ  —  C  the  quantity  of  water  poured  out,  C  being  the  tipped 
edge  of  the  bottom. 

I.  A  tank  is  6  feet  in  diameter  at  the  top,  8  feet  at  the  bot- 
tom, and  12  feet  deep.  A  plane  passes  from  the  top  on  one  side 
to  the  bottom  on  the  other  side  :  into  what  segments  does  it 
divide  the  tank? 

By  formil,a,  K= 
327r(8-3V/3)=281.87  cu  ft. 


II. 


1.  4  ft.=  AL,  the  radius  of  the  lower  base. 

2.  3  ft.=Z?/?1,  the  radius  of  t"he  upper  base,  and   • 

3.  12  ft.=^Z,  the  altitude.     Then 

_V3"8)=  327r(8-3\/3)=281.87 


3(4-3) 
sq.  ft.— the  volume. 

III.     .'.  The  volume  is  281.87  cu.  ft 

Prob.  XCVIII.  To  find  the  convex  surface  ofa  conical  un- 
gula,  when  the  cutting-  plane  FCE  makes  an  angle  CIB  less 
than  the  angle  DAB,  i.  e.  when  AIi=t)  is  less  thanDC(=2r). 


Formula.— S= 


This  formula  is  obtained  by  integrating  the  formula  of 
Prob.  XCIV,  recollecting  that  the  co-  efficient  of x2  is  negative. 

Prob.  XCIX.  To  find  the  volume  of  a  conical  ungrula,  when 
the  cutting' plane  FCE  makes  an  angle  CIB  less  than  the  an- 
gle DAB.  i.  e.,  when  AI  (=t)  is  less  than  CD  (^3r). 


MENSURATION. 


269 


Formula.-  V= 


FIG.  44. 


This  formula  is  obtained  by  integrating  the 
formula  of  Prob.  XCV,  recollecting  that  the  co- 
efficient of  x2  is  negative. 

Prob.  C.  To  find  the  convex  surface  of  a  conical  un- 
gula,  when  the  cutting-  plane  FCE  is  parallel  to  the  side  AD, 
i.  e.,  when  AI(=t)  is  equal  to  DC(=2r). 


Formula.—  S== 


cos 


-r)r-%(R-r^(R—  r}r\ 


This  formula  is  obtained  by  putting  /=2r, 
in  the  formula  of  Prob.  XCIV.,  and  inte- 
grating the  resulting  equation. 

Fob.  CI.  To  find  the  volume  of  a  coni- 
cal uiig-ula,  when  the  cutting-  plane  FCE 
is  parallel  to  the  side  DA,  i.  e.,  when 
AI(=t)  is  equal  to  CD  (=2r). 


Formula.-  V 


FIG   45. 


This  formula  is  obtained  by  putting  /=2r,  in  the  formula  of 
Prob.  XCV.,  and  integrating  the  resulting  equation. 

Prob.  CII.  To  find  the  con  vex  surface  of  a  concial  ung-ula, 
when  the  cutting  plane  FCE  makes  an  angle  C1B  greater 
than  the  angle  DAB,  i.  e.,  when  AI  (=t)  is  greater  than  DC 


Formula.- 


270  FINKEL'S  SOLUTION  BOOK. 

This  formula  is  obtained  by  integrating  the  formula  of  Prob. 
XCIV.,  remembering  that  the  coefficient  of  x2,  which  occurs  in 
process  of  integrating,  is  positive. 

Prob.  CHI.  To  find  the  volume  of  a  conical  nngula,  when 
the  cutting  plane  FCE  makes  an  angle  CIB  greater  than  the 
angle  DAB,  i.  e.,  when  AI(=t)  is  greater  than  DC(=2r). 


Formula 


This  formula  is  obtained  by    integrating   the    formula  of  Prob, 
XCV.,  regarding  the  coefficient  of*2  positive. 


XII.  THE 
Prob.  CIV.    To  find  the  convex  surface  of  a  sphere. 


Formula.—  S=2  X  2^yV^>'34-^2=47r7?2=  ?rZ>2,    where 
Z?  is  the  diameter. 

Rule.  —  Multiply  the  square  of  the  diameter  by  3.14-1592. 
I.     What  is  the  surface  of  a  sphere  whose  radius  is  5  inches? 
By  formula,  S=±7tR  3=4^X25=314.1592  sq.  in. 

II.  5  in.=the  radius. 
2.  25  sq.  in.=  the  square  of  the  radius. 
3.  /.  4#X25  sq.  in.=314.1592  sq.  in.=the   surface   of  the 
^  sphere. 

III.     .'.  314.1592  sq.  in.^the  surface  of  the  sphere. 

NOTE.  —  Since  7r/?2is  the  area  of  a  circle  whose  radius  Is  /?,  the  area 
(4n-/?2)  of  a  sphere  is  equal  to  four  great  circles  of  the  sphere.  The  sur- 
face of  a  sphere  is  also  equal  to  the  convex  surface  of  its  circumscribing 
cylinder. 

Proh.  C  V.    To  find  the  volume  of  a  sphere,  or  a  globe. 
Formula.—  V=&ity*  '<**=£*•  R*=  f  x(\D)  3=| 


Rule.  —  Multiply  the  cube  of  the  radius  by  far  (=4.188782);  or 
multiply  the  cube  of  the  diameter  by  \TT  (=.5235987). 

I.      What  is  the  volume  of  a    sphere    whose  diameter  is  4  feet? 


MENSURATION.  271 

By  formula,  F=f7T./?3=|7r23=33.510256cu.  ft. 

1.  2  ft.=the  radius. 

2.  8  cu.  ft=23=the  cube  of  the  radius. 

t<!3.  /.  4.188782X8  cu.  ft.=33.510256  cu.  ft.==the   volume   of 

the  sphere. 

III.     .-.  33.510256  cu.  ft.=the  volume  of  the  sphere. 
Prob.  CVI.    To  find  the  area  of  a  zone. 

A.  ZiOn,e  Is  the  curved  surface  of  a  sphere  included  between  two 
parallel  planes  or  cut  off  by  one  plane. 

Formula. — S=^nRa,  in  which  a  is  the    altitude  of  the 
segment  of  which  the  zone  is  the  curved  surface. 

Rule. — Multiply  the  circumference  of  a  great  circle  of  the 
sphere  by  the  altitude  of  the  segment. 

I.  What  is  the  area  of  a  zone  whose  altitude  is  2  feet,  on  a 
sphere  whose  radius  is  6  feet? 

By  formula,  5=2*^=2  ?rQx 2=24 n  =75.39822  sq.  ft. 

1.  6  ft.=the  radius  of  the  sphere. 

2.  2  ft.=the  altitude. 

II.<(3.  12  ?r=37. 69911  ft.=the  circumference  of  a  great  circle  of 

the  sphere. 

4.  .'.2X37.69911=75.39822  sq.  ft.=the  area  of  the  zone. 
III.     .-.  The  area  of  the  zone  is  75.39822  sq.  ft. 

NOTE. — This  rule  is  applicable  whether  the  zone  is  the  curved  surface  of 
the  frustum  of  a  sphere  or  the  curved  surface  of  a  segment  of  a  sphere. 

Prob.  CVII.    To  find  the  volume  of  the  segment  of  a  sphere. 

Formula. —  V=.\na(s&r\-\-aCL)  where  r1  is  the   radius  of 
the  base  of  the  segment. 

Rule. —  To  three  times  the  square  of  the  radius  of  the  base,  add 
the  square  of  the  altitude  and  multiply  the  sum  by  ^n=.52S5987 
times  the  altitude. 

I.  What  is  the  volume  of  a  segment  whose  altitude  is  2  inches 
and  the  radius  of  the  base  8  inches? 

By  formula,  F=l7ra(3r;+*2)=-^X2(3X64+4)=205.2406 
cu.  in. 

1.  8  in.=the  radius  of  the  base. 

2.  2  in.=the  altitude  of  the  segment. 

3.  192  sq.  in  =g3x82=three  times  the  square  of  the  radius. 


II. 


4.  4  sq.  in.=the  square  of  the  altitude. 

5.  196  sq.  in.=192  sq.  in.-f4  sq.  in. =three  times  the  square 

of  the  radius  plus  the  square  of  the  altitude. 

6.  |7TX2 X  196=205.2406  cu.  in.=the  volume  of  the  segment- 


272  FINKEL'S  SOLUTION  BOOK. 

III.      /.  205.2406  cu.  in  =the  volume  of  the  segment. 


NOTE.  —  From  the  formula  F=^a(3r-(-rt2)>  we  nave  V—  \-xa 
But  \xar\  is  the  volume  of  a  cylinder  whose  radius  i$rlt  and  altitude  {a,  and 
i7r«3  is  the  volume  of  a  sphere  whose  diameter  is  a  :.  The  volume  of  a 
segment  of  a  sphere  is  equal  to  a  cylinder  whose  base  is  the  base  of  the  seg- 
ment and  altitude  half  the  altitude  of  the  segment,  plus  a  sphere  whose 
diameter  is  the  altitude  of  the  segment. 

Prob.  CVIII.  To  find  the  volume  of  a  frustum  of  a 
sphere  ,or  the  portion  included  between  two  parallel  planes. 

Formula.—  V^7ta\Z(rl+rl)+a*-\=\a(nrl+nr\)  + 
\7t  a*  *,  in  which  rl  is  the  radius  of  the  lower  base,  r2  the  ra- 
dius of  the  upper  base. 

Rule.  —  To  three  times  the  sum  of  the  squared  radii  of  the  two 
ends,  add  the  square  of  the  altitude;  multiply  this  sum  by  .5235987 
times  the  altitude. 

I.  What  is  the  volume  of  the  frustum  of  a  sphere,  the  radius  of 
whose  upper  base  is  2  feet  and  lower  base  3  feet  and  altitude  -J- 
foot? 

By  formula,    V=\  *a[3(rH-rJ)+a*]=i,r  xi[3(9+4)+i]= 
8.03839  cu.  ft. 

1.  3  ft.—  the  radius  of  the  lower  base. 

2.  2  ft.—  the  radius  of  the  upper  base. 

3.  39  sq.  ft.=--3(32-f22)=three  times  the  sum  of  the  squares 


II. 


of  the  radii  of  the  two  bases. 


4.  \  sq.  ft.=the  square  of  the  altitude. 

5.  .-.  |TT  XiX39^=8.03839  cu.    ft=the     volume     of     the 

frustum. 

III.     .-.  8.03839  cu.  ft.=the  volume  of  the  frustum. 
Prob.  CIX.    To  find  the  volume  of  spherical  sector. 

A  Spherical  Sector  is  the  volume  generated  by  any  sector 
of  a  semi-circle  which  is  revolved  about  its  diameter. 

Formula. —  l/=^7ta7?2 ,  where  a   is  the    altitude  of  the 
zone  of  the  sector. 

Rule. — Multiply  its  zone  by  one-third  the  radius. 

*NOTK. —  £rt(?r ;•?-(- Trr j )=the  volume  of  two  cylinders  whose  bases  are 
the  upper  and  lower  bases  of  the  segment  and  whose  altitude  is  hall  the  alti- 
tude oi  the  segment.  ^Tra3  is  the  volume  of  a  sphere  whose  diameter  is  the 
altitude  of  the  segment.  Hence  the  volume  of  a  segment  of  a  sphere  of  two 
bases  is  equivalent  to  the  volume  of  two  cylinders  whose  bases  are  the  up- 
per and  lower  bases  respectively  of  the  segment  and  whose  common  altitude 
is  the  altitude  of  the  segment,  plus  the  volume  of  a  sphere  whose  diameter 
is  the  altitude  of  the  segment. 

For  a  demonstration  of  this  and  the  preceding  formula,  see  Wenttvorth's 
Plane  and  Solid  Geometry,  Bk.  IX.,  Prob.  XXXII. 


MENSURATION. 


273 


I.     What  is  the  volume  of  a    spherical    sector   the   altitude  of 
whose  zone  is  2  meters  and  the  radius  of  the  sphere  6  meters? 

By  formula,  F=f  jw/?a=fjr  X2x62= 
150.7964m3. 

"1.  2m.=the  altitude  BD  of  the  zone  gener- 
ated by  the  arc  EF  when  the  semicir- 
cle is  revolved  about  AB. 

2.  6m.=the  radius  EC  of  the  sphere. 

3.  2  7r6m.=37. 699104  m  =the  circumference 

of  a  great  circle  of  the  sphere. 

4.  2 7t 6 X 2=75.398208  m2.=the   area   of  the 

zone  generated  by  EF,  by  Prob.  CVI. 

5.  .-.  ix6x75.398208=150.796416ms=the     volume  of  the 

spherical  sector. 

Ill      .-.  The  volume  of  the  spherical  sector  is  150.796416  m8. 

I.     Find  the  diameter  of  a  sphere  of  which    a    sector  contains 
7853.98  cu.  ft.,  when  the  altitude  of  its  zone  is  6  feet. 


II., 


FIG.  47. 


By  formula,   V= 
7853.98  cu.  ft,    or   4r2=2500  sq.  ft.,    whence  2r=50   feet,    the 
diameter  of  the  sphere. 

1.  6  ft.=the  altitude  of  the  zone. 

2.  .'.  \ir.  x6X?'2=the  volume  of  the  sector.     But 

3.  7853.98  cu.  ft=the  volume. 

4.  .-.  |7rx6X^2==7853.98  cu.  ft. 

5.  r2=625  sq.  ft,  by  dividing  by  4;r. 

6.  .'.  2r=50  ft.,  the  diameter  of  the  sphere. 

III.     .-.  The  diameter  of  the  sphere  is  50  feet. 
Prob.  CX.    To  find  the  area  of  a  lune. 


II. 


A.  Lime  is  that  portion  of  a   sphere    comprised    between  two 
great  semi-circles. 

*  f  A 

Formula.  —  S=^7tR^{    — 

the  quotient  of  the  angle  of  the  lune  divided  by  360°. 


where     u    is 


.  —  Multiply  the  surface  of  the  sphere  by   the   quotient  of 
the  angle  of  the  lune  divided  by  360° 

I.     Given  the  radius  of  a  sphere  10  inches;  find  the   area   of  a 
lune  whose  angle  is  30°. 


By  formula,  S=±7t  R*  u=4X  *  XlO2  X(30°-f-360°  )= 
£7rl02=104.7197  sq.  in. 


274  FINKEL'S  SOLUTION  BOOK. 

1.  10  in.=the  radius  of  the  sphere. 

2.  30°=the  angle  of  the  lune. 

3.  T1Tp=30°-r-360°=:the  quotient  of  the    angle   of  the  lune 


II. 


divided  by  360°. 

4.  47rl02=400^=1256.6368  sq.   in.=the    suiface    of  the 

sphere. 

5.  .'.  TVX  1256.6368  sq.  in.=104.7198  sq.  in.=the  area  of  the 

lune. 
III.     .-.  The  area  of  the  lune  is  104.7198  sq.  in. 

Went-wortW  s  New  Plane  and  Solid  Geometry,  p.  371,  Ex.  585. 
Prob.  CXI.    To  find  the  volume  of  a  spherical  ungula. 


A.  Spherical  U^ngula  is  a  portion  of  a  sphere  bunded  by  a 
lune  and  two  great  semi-circles. 

Formula.  —  V=\n  R*u,  where  u  is    the  same  as  in  the 
last  problem. 

Rule.  —  Multiply  the  area  of  the  lune  by  one-third  the  radius; 
or,  multiply  the  volume  of  the  sphere  by  the  quotient  of  the  angle 
of  the  lune  divided  by  360°. 

I.  What  is  the  volume  of  a  spherical  ungula  the  angle  of 
whose  lune  is  20°,  if  the  radius  of  the  sphere  is  3  feet? 


By    formula,     V=$  *  R*  U=±TT  x  33  X(20°-r-360°)  =  6.283184 
cu.  ft. 


II. 


rl.  3  ft.=the  radius  of  the  sphere. 

2.  47r32X(20°-j-360°>=6.2S3184sq.  ft.=the   area   of  the 
lune,  by  Prob   CX 


3.  .'.  £X3X6.283184=6.283184  cu.  ft.=the  volume   of  the 

ungula. 
III.     .-.  6.283184  cu.  ft.  is  the  volume  of  the  Ungula. 

Prob.  CXII.    To  find  the  area  of  a  spherical  triangle. 

Formula.— 3=2* R*  x  (A^-B+C—  180°)-r-360°,  in 
which  A,  B,  and  C  are  the  angles  of  the  spherical  triangle. 

Rule. — Multiply  the  area  of  the  hemisphere  in  which  the  tri- 
angle is  situated  by  the  quotient  of  the  spherical  excess  (the ex- 
cess of  the  sum  of  the  spherical  angles  over  180^)  divided  by  360°. 

I.  What  is  the  area  of  a  spherical  triangle  on  a  sphere 
whose  diameter  is  12,  the  angles  of  the  triangle  being  82°,  98°, 
and  100°  ? 

By  formula,  5=2  nR*  x(A+£-\-C—  lSO°)-r-3600=27r62  X 
(82°+980+100°— 


MENSURATION.  275 

1.  6=the  radius  of  the  sphere. 

2.  27r62=727r=the  area  of  the  hemisphere. 

T  ,3.   (82°+98°+1000— 180°)=100°=the  spherical  excess. 

4.  100 ° -7-360  °=T5¥=the  quotient  of   the  spherical  excess 

divided  by  360°. 

5.  .'.  T53X72nr=*62.83184=the  area  of  the  spherical  triangle. 
III.     .'.  The  area  of  the  spherical  triangle  is  62.83184. 
(Olney*s  Geometry  and  Trigonometry,  Un.  Ed., p .238, Ex.  8.) 

Prob.  CXIII.    To  find  the  volume  of  a  spherical  pyramid. 

A  Spherical  Pyramid  is  the  portion  of  a  sphere  bounded 
by  a  spherical  polygon  and  the  planes  of  its  sides. 

Formula.—  F=f7r/?8X(  ^-5-360°),    where    B  is    the 
spherical  excess. 

Rule. — -Multiply  the  area  of  the  base  by  one-third  of  the  radius 
of  the  sphere 

I.  The  angles  of  a  triangle,  on  a  sphere  whose  radius  is  9  feet, 
are  100°,  115°,  and  120°  ;  find  the  area  of  the  triangle  and 
the  volume  of  the  corresponding  spherical  pyramid. 

By  formula,     F=| n R*  x  ( .£-7-360 °  }=\nR*  K(A+B+C— 
18G0)-7-360°  =  f  nr93X  (100° +115° +120°— 180°  ) -j- 360°  = 
'»657.a771$6  cu.  ft. 

1.  9  ft.=the  radius  of  the  sphere. 

2.  27r92=the  area  of  the  hemisphere  in  which  the  pyramid 

is  situated. 

3    (100° +115° +120°— 180°)— 155°=the    sperical    ex- 
cess. 


II. 


4.  fi=l 55° -r-36G°=the  quotient  of  the    spherical    excess 


divided  by  360°. 
5.  .-.  fix27r92HBX  ^92=the  area  of  the  base  of  the  pyra- 

mid. 
•-  iX9XliX27r92=657.377126cu.    ft.=the    volume    of 

the  pyramid. 

III.  .'.  The  volume  of  the  spherical  pyramid  is  657.377126 
cu.  ft. 

(  Van  Amringe's    Da-vies*    Geometry  and  Trigonometry,  p.  278, 
Ex.  15. 

I.  Fmd  the  area  of  a  spherical  hexagon  whose  angles  are  96  °  , 
110°,  128°,  136°,  140°,  and  150°,  if  the  circumference  of  a 
great  circle  of  the  sphere  is  10  inches. 

Formula.—  S=<l7tR*  [  T~~(  U~™  °  \    where    T  is 


the  sum  of   the    angles    of  the    polygon   and    n    the    number    of 
sides. 


276  FINKEL'S  SOLUTION  BOOK. 


By  formula, 

(96° +110° +128° +136° +140° +150°—  (6— 2)xl80°)-5- 
360°  =  —  X(760°—  720°  )^-360°=i— =1.7684  sq.  in. 

7t  7t 

1.  5-7-'T=the  radius  of  the  sphere,  since  2?rR— 10  in.. 

2.  760°=960+1100+1280+1360+1400+150°=the 

sum  of  the  angles  of  the  polygon. 

3.  760°—  (6— 2)xl80°=40°=the  spherical  excess. 

4.  ^=400-r-360°=the  quotient  of  the   spherical  excess  di- 
rr<  videdby360°. 

><5  X2 

5.  2^1—1  —the  area  of  the   hemisphere    on    which    the 

polygon  is  situated. 

(5  \2 
-   1  =4x50-f-JT=1.7684  sq.  in. 
7t  S 

III.     /.  The  area  of  the  polygon  is  1.7684  sq.  in. 
Wentworttts  Geometry,  Revised  Ed., p.  374,  Ex.  596. 


XIII.  SPHENOID. 

1.  A  Spheroid  is  a  solid  formed  by  revolving  an  ellipse 
about  one  of  its  diameters  as  an  axis  of  revolution. 

1.  THE  PROLATE  SPHEROID. 

1.  The  Prolate  Spheroid  is  the  spheroid  formed  by  re- 
volving an  ellipse  about  its  transverse  diameter  as  an  axis  of 
revolution. 

Prob.  CXIV.    To  find  the  surface  of  a  prolate  spheroid. 

Formulae.— (a)  S= 


,  where 
&  & 

^2Z_32 


a 
surface. 


=the  eccentricity  of  the   ellipse  which  generates    the 


--— 


MENSURATION. 


277 


Rule. — Multiply  the  circumference  of  a  circle  whose  radius 
$s  the  semi -conjugate  diameter  by  the  semi -conjugate  diameter  in- 
creased by  the  product  of  the  arc  whose  sine  is  the  eccentricity  into 
the  quotient  of  the  semi-transverse  diameter  divided  by  the  eccen- 
tricity. 

I.  Find  the  surface  of  a  prolate  spheroid  whose  transverse 
diameter  is  10  feet  and  conjugate  diameter  8  feet. 

5    . 


By  formula  (a),  S=27rt>(l>-\- 

e 


rA 


|7r[48+100x.  6435053]  —235.3064  sq.  ft. 

fl.  25.  1327412=2  7r4=the    circumference    of  a  circle  whose 
radius  is  the  semi-conjugate  diameter  of  the  ellipse. 


II.  < 


2.     = 


=the  eccentricity. 


3.  2^5  ft.=5  ft-i-f =the  quotient  of  the  semi-transverse  diame- 

ter divided  by  the  eccentricity. 

4.  .6435053=the  arc  (to  the  radius  1)  whose  sine  is  f ,  or  the 

eccentricity. 

5.  5.3625442  ft.=Yft.  X  .6435053= %5  ft.  X  the  arc  whose 

sine  is  -| 

6.  9.3625442  ft.=4  ft+5.3625442  ft=semi  -  conjugate    di- 

ameter increased  by  said  product. 

7.  .-.  235.3064  sq.  ft.=9.3625442x25.1327412=the   surface 

of  the  prolate  spheroid. 

III.     /.  The  surface  of  the  prolate  spheroid  is  235.3064  sq.  ft. 
Prob.  CXV.    To  find  the  volume  of  a  prolate  spheroid. 

1)  2    /*a 

Formula.—  F=  Cny*dx=7t—  I     (a2—x2)dx= 
J  a1J _a 

r—  I     a2x — £#3          =^7rb'2a,  in  which    b    is   the    semi-conjugate 

I—  I  —  d 

diameter,  and  a  the  semi-transverse  diameter. 

Rule. — Multiply  the  square  of  the  semi -conjugate  diameter  by 
the  semi-transverse  diameter  and  this  product  by  ^n. 

I.     What  is  the  volume  of  a  prolate  spheroid,  whose  semi-trans- 
verse diameter  is  50  inches,  and  semi-conjugate  diameter  30 inches. 

By  formula,    r=|^2«=|^302  X 50=188495.559  cu.  in. 


278  FINKEL'S  SOLUTION  BOOK. 

1.  30  in.=the  semi -conjugate  diameter, 

2.  50  in.=the  semi-transverse  diameter. 

3.  900  sq.  in=the  square  of  the  semi-conjugate  diameter. 

[.  45000  cu.  in.=50x900=the  squareof  the  semi-conjugate 

diameter  by  the  semi- transverse  diameter. 
5.  . -4^45000=1X3.14159265X45000 cu.  in.= 

188495.559  cu.  in.=the  volume  of  the  prolate  spheroid. 
III.     .*.  The  volume  of  the  prolate  spheroid  is  188495.559  cu.  in. 

2.  THE  OBLATE  SPHEROID. 

1.  An  Oblate  Spheroid  is  the  spheroid  formed  by  revolving 
an  ellipse  about  its  conjugate  diameter  as  an  axis  of  revolution. 

Prob.  CXVI.    To  find  the  surfae  of  an  oblate  spheroid. 

ra 

Formulae. — (a)  S=  C27rxds=2  I    2; 

•'-a 


q 


Prob.  CX  VII.    To  find  the  volume  of  an  oblate  spheroid. 


II  l  lie.  —  Multiply  the  square  of  the  semi-transverse    diameter- 
by  the  semi.  conjugate  diameter  and  this  product  by  ^  n. 

I.     What  is  the  volume  of  an    oblate    spheroid,    whose    trans- 
verse diameter  is  100  and  conjugate  diameter  60? 


By  formula,   V=$7t  a*  6=$ie5Q2  X30=314159.265. 

1.  30=^  of  60=  the  semi-conjugate  diameter. 

2.  50—  -J  of  100=the  semi-transverse  diameter. 

3.  2500=502=the  square  of  the  semi-transverse  diameter. 


II. 


4.  75000=30  X2500=the  square  of   the  semi -transverse  di- 


ameter multiplied  by  the  semi-conjugate  diameter. 

5.  /.  \n  X 75000=314159.265=  the    volume  of   the  oblate 

spheroid. 
III.     .-.  The  volume  of  the  oblate  spheroid  is  314159.265. 

NOTE  — Since  the  volume  of  a  prolate  spheroid  is  ^b^a.  We  may  write 
§7r£2a— §(7r£2X2rt).  But  7r^2x2«  is  the  volume  of  a  cylinder  the  radius  of 
whose  base  is  b  and  altitude  2a.  .'.  The  volume  of  a  prolate  spheroid  is  |  of 
the  circumscribed  cylinder.  In  like  manner,  it  may  be  shown  that  the  vol- 
ume of  an  oblate  spheroid  is  §  of  its  circumscribed  cylinder. 

The  following  is  a  general  rule  for  finding  the  volume  of  a 
spheroid;  Multiply  the  square  of  the  revolving  axis  by  the  Jixed 
axis  and  this  product  by  \it . 


MENSURATION. 


Prob.  CXVIII.  To  find  the  volume  of  the  middle  frustum 
of  a  prolate  spheroid,  its  length,  the  middle  diameter,  and 
that  of  either  of  the  ends  being  given. 

CASE  I. 

When  the  ends  are  circular,  or  parallel   to    the  revolving  axis. 
F=TVn'(2Z>2+^2)/,    where    D    is    the    middle 


diameter  CD,  d  the  diameter  ///of  an    end,  and  /  the  length  of 
the  frustum. 

Rule.  —  To  twice  the  square  of  the  middle  diameter  add  the 
square  of  the  diameter  of  either  end  and  this  sum  multiplied  by 
the  length  of  the  frustum,  and  the  product  again  by  -±^n,  'will 
give  the  solidity. 

I.  What  is  the  volume   of  the   middle  frustum    HIGF  of  a 
prolate  spheroid,  if  the  middle  diameter  CD   is    50   inches,    and 
that  of  either  of  the  ends  ///or  FG  is  40    inches,   and  its  length 
OK  18  inches? 

By  formula,   F=T1^7r(2/?2+c/2)/=T1Tr7r(2x502+402)18= 
31101.767265  cu.  in. 

1.  50  in.=the  middle   diame- 

ter CD. 

2.  40  in.=the    diameter  of  ei- 

ther end  as  HI. 

3.  18  in.=the   length    OK  of 

the  frustum. 

4.  5000sq.  in.=2x502=twice 

II.  <  the  square  of  the  middle 

diameter. 

5.  1600sq.  in.=402=the 

sqaureof  the  diameter  of  either  end. 

6.  5000  sq.  in.+1600  sq.  in.=6GOO  t.q.  in. 

7.  18X6600=118800  cu.  in. 

8.  .'.  TVwXH8800cu.  in.=31101.767265  cu.    in.=the  vol- 

ume. 

III.  .-.  The  volume  of  the  frustum  is  31101.767265  cu.  in. 

CASE  II. 

When  the  ends  are  elliptical,  or  perpendicular  to  the  revolving 
axis. 


.—  V=^n(ZDd+D'd')l,  where  D  and  d  are 
the  transverse  and  conjugate  diameters  of  the  middle  section  and 
D/  and  d/  the  transverse  and  conjugate  diameter  of  the  ends  and 
/  the  distance  between  the  ends. 

Rule.  —  (  1  )  Multiply  twice   the    trans-verse   diameter  of  the 
middle  section  by  its  conjugate  diameter,  and  to  this  product  add 


280  FINKEL'S  SOLUTION  BOOK. 

the  product  of  the  transverse  and  conjugate  diameter   of  either  of 
the  ends. 

(2)  Multiply   the  sum,  thus  found,  by  the  distance  of  the  ends, 
the  height  of  the  frustum, 
the  result  will  be  the  volume. 


or  the  height  of  the  frustum,  and  the  product  again   by  ^7t 
the 


I.  What  is  the  volume  of  the  middle  frustum  of  an  oblate 
spheroid,  the  diameter  of  the  middle  section  being  100  inches 
and  60  inches;  those  of  the  end  60  inches  and  36  inches  ;  and  the 
length  80  inches? 

By  formula,  V=^7t(%Dd-\-D'd')  /=^w(2x  100x60+60 X 
36)  80=296566.44616  cu.  in. 

1.  100  in.=the  transverse  diameter  PC  of  the  middle  section. 

2.  60  in. =the  conjugate  diameter  ms  of  the  middle  section. 

3.  12000sq.  in.==2XlOOX60==twice     the   product    of  the 

diameters  of  the  middle  section. 

4.  60  in.=the      transverse     diameter 

AB  of  the  end. 

5.  36  in.=the     conjugate    diameter 
•r-r  J  %(nc)  of  the  end. 

6.  2160  sq.  in.=the   product   of  the 

diameters  of  the  end. 

7.  14160  sq.  in.=12000  sq.  in.+2160 

sq.  in. 

8.  80X14160=1132800  cu.  in.  =  the  FIG.  49. 

product  of  said  sum  by  the  height  of  the  frustum. 

9.  .'.  TV*X  1132800  cu.  in.=296566.44616  cu.   in.=the  vol- 

ume of  the  frustum. 

III.     .-.  The  volume  of  the  frustum  is  296566.44616  cu.  in. 

Prob.  CXIX.  To  find  the  volume  of  a  segment  of  a  prolate 
spheroid 

CASE  I. 
When  the  base  is  parallel  to  the  revolving  axis. 

Formula.— V=\nh*(--\' (ZD—Wi],  where  h  is   the 

height  of  the  segment,  d  the    revolving    axis,    and   D    the    fixed 
axis. 

Rule. — (1)  Divide  the  square  of  the  revolving  axis  by  the 
square  of  the  fixed  axis,  and  multiply  the  quotient  by  the  differ- 
ence  between  three  times  the  Jixed  axis  and  twice  the  heigat  of 
the  segment. 

(2)  Multiply  the  product,  thus  found,  "by  the  square  of  the 
height  of  the  segment,  and  this  product  by  \TT,  and  the  result  will 
be  the  volume  of  the  segment. 


MENSURATION.  281 

I.  What  is  the  volume  of  a  segment  of  a  prolate  spheroid  of 
which  the  fixed  axis  is  10  feet  and  the  revolving  axis  6  feet  and 
the  height  of  the  segment  1  foot? 


By  formula,   V=\nh*  (%D—  2h)= 


5.277875652  cu.  ft. 

1.  10  ft.=the     transverse      diameter 

1BF.  0 

2.  6  ft.=the  conjugate  diameter  AE. 

62  ' A 

3.  ¥9^=  —  =the  square  of  the  conju- 


II. 


gate    diameter   divided     by   the 
square  of  the  transverse  diameter. 

4.  28ft.=3XlO  ft— 2X1  ft.=  the  difference  between  three 

times  the  transverse  diameter  and  twice    the  height  of 
the  segment. 

5.  /7X28  ft.=10&  ft.=the  product   of  said    quotient   by 

said  difference. 

7.  .-.  iVxlOA- cu.  ft.=5.277865652  cu.  ft.=the  volume. 


III.     /.  The  volume  of  the  segment  is  5.277875652  cu.  ft. 

CASE  II. 
When  the  base  is  perpendicular  to  the  revolving  axis. 

Formula.—  V=±7r/i2(--.J(3d—  <2k),  where  d  is    the 
revolving  axis,  D  the  fixed  axis,  and  h  the  height  of  the  segment. 

Rule. — (1)  Divide  the  fixed  axis  by  the  revolving  axis,  and 
multiply  the  quotient  by  the  difference  between  three  times  the 
revolving  axis  and  twice  the  height  of  the  segment, 

(2).  Multiply  the  product,  thus  found,  by  the  square  of  the 
height  of  the  segment,  and  this  product  again  by  \rc. 

I.  Required  the  volume  of  the  segment  of  a  prolate  spheroid, 
its  height  being  6  inches,  and  the  axes  50  and  30  inches  respect- 
ively. 

By  formula,   V<^tk*(£)  (3<*-2>&)=^x62  (|J)x 


282  FINKEL'S  SOLUTION  BOOK 

,(3X30— 2X6)=2450.442267  cu.  in. 

1.  50  in.=the    transverse    diameter,  or 

axis. 

2.  30  in.=the  conjugate  diameter  2MO. 

3.  !=50-=-30=the  quotient  of  the  trans 

verse    diameter     divided    by    the 

conjugate  diameter.  FIG.  57. 

4.  78  in.=3x30  in. — 2x6in.=the  difference  between  three 

times  the  conjugate  or  revolving    axis,   and  twice  the 
height  of  the  segment. 

5.  130in.=|x78  in.=the  product  of  said  quotient  by  said 

difference. 

6.  4680  cu.  in=130x62=the  square  of  the    height    of   the 

segment  by  said  product. 

7.  .'.  £tfX4680  cu.  in.=2450.442269  cu.  in.=the  volume  ot 

segment. 

.III.     .-.  The  volume  of  the  segment  is  2450.442269  cu.  in. 

XIV.  CONOIDS. 

1.     A  Conoid  is  a  solid  formed  by  the  revolution  of  a  conic 
section  about  its  axis. 

I.  THE  PARABOLIC  CONOID. 

1.  A  Parabolic  Conoid  is  the  solid  formed  by  Devolving  a 
parabola  about  its  axis  of  abscissa. 

Prob.  CXX.    To  find  the  surface  of  a  parabolic  conoid,  or 
paraboloid. 

Formulae. — (a)  S— 


where  2/  is  the  latus  rectum  of  the  parabola    and y  the    radius  of 
the  base  of  the  conoid,  or  the  ordinate  of  the  parabola. 

(3)   5— fTTV^j  (p-\-*)*— p*\j  where  2^   is   the    same    as    above 

and  x  the  altitude  of  the  conoid,  or    the    axis   of  abscissa    of  the 
parabola. 

Rule. —  To  the  square  of  half  the  latus  rectum,  or  principal 
parameter,  add  the  square  of  the  radius  of  the  base  of  the  conoid 
and  extract  the  square  root  of  the  cube  of  the  sum;  from  this  re- 
sult, subtract  the  cube  of  half  the  latus  rectum  and  multiply  the 


MENSURATION. 


difference  ~by2n,  and  divide  the  product  by  one  and  one-  half  times 
the  latus  rectum. 

I.     Determine  the  convex  surface  of  a  paraboloid  whose  axis  is 
20,  and  the  diameter  of  whose  base  is  60. 

From  the  equation  of  the  parabola,  yz= 
we  have302=2/x20;  whence  2/=45. 

•.  By 


|n-X25x(125—  27)=49X25X3.14159265= 
3848.45118. 


FIG.  52. 


II. 


1.  30=  the  radius  AO  of  the  base  of  the  conoid. 

2.  20=the  altitude  OD.     Then  by  a  property   of  the  para- 

bola, 

3.  302=2/X  20,  whence 

4.  p=22$,  the  principal  parameter  of  the  parabola. 

5. 


=the  square   root  of 

the  cube  of  the  sum  of  the  squares  of  half  the    latus 
rectum  and  the  radius  of  the  base. 

6.   1    —  1  =the  cube  of  half  the  latus  rectum.     . 


ence  between  said  square  root  and    the    cube   of   half 
the  latus  rectum. 


o      n      vx  •  m     /  -|  o  c o>7  \ TrN/QSVl     ~"     I    ^TT     timPS 

said  difference. 
9.  .'.  2^98x(-^->)  -HlX45)=3848.45118=the  surface  of 

the  conoid. 

III.     .-.  The  surface  of  the  conoid  is  3848.45118. 
Prob.  CXXI.    To  find  the  volume  of  a  parabolic  conoid. 


altitude. 


Formula.  —  V=Jrty'*dx=j7tflp  x 
)x—%7ry2x,  whereby  is  the  radius  of 


the   base    and    x  the 


Rule.  —  Multiply  the  area  of  the  base  by   the  altitude  and  take 
half  the  product. 


284  FINKEL'S  SOLUTION  BOOK. 

T.  What  is  the  volume  of  parabolic  conoid,  the  radius  of 
whose  base  is  10  feet  and  the  altitude  14  feet? 

By  formula,  V=\n  y*  x  =  %n\W  X  14=700x^=2202.  114855 
cu.  ft. 

1,  10  ft.=the  radius  of  the  base. 

2.  14  ft.^the  altitude. 

II.<3.   7rl02=314.159265sq.  ft.  the  area  of  the  base. 

(       /.  |X14X314.159265=-2202.114855  cu.  ft.=the   volume 
of  the  conoid. 

III.     .-.  The  volume  of  the  conoid=2202.114855  cu.  ft. 
NOTE.  —  Since  the  volume  of  the  conoid  is  %Ky2x,  it  is  half  of  its   circum- 
scribed cylinder. 

Prob.  CXXII.  To  find  the  convex  surface  of  a  frustum  of 
a  parabolic  conoid  of  which  the  radius  of  the  lower  base  is  R 
and  the  upper  base  r. 

In  (  3  a 

Formula.  <  ti22 


rR  n 

.—  S=J   <2l7tyds=jj) 


I.  What  is  the  volume  of  the  frustum  of  a  parabolic  conoid  of 
which  the  radius  of  the  lower  base  is  12  feet,  the  radius  of  the 
upper  base  8  feet,  and  the  altitude  of  the  frustum  5  feet? 

Since  122=2/*/  and  82=2px,  122—Sz=2p(x/—x).  Bnt*'—  x 
=5  feet.  /.  122—  82=2/x5,  whence  2/=16,  the  latus  rectum. 

.-.  By  formula,  sJ- 


Prob.  CXXIII.  To  find  the  volume  of  the  frustum  of  a 
parabolic  conoid,  when  the  bases  are  perpendicular  to  the 
axis  of  abscissa. 

Formula.—  V=^7tRzx/—\7rr^x=^7i(x/ 


Rule.  —  Multiply  the  sum  of  the  squares  of  the  radii  of  the 
two  bases  by  n  and  this  product  by  half  the  altitude. 

I.  What  is  the  volume  of  the  frustum  of  a  parabolic  conoid, 
the  diameter  of  the  greater  end  being  60  feet,  and  that  of  the 
lesser  end  48  feet,  and  the  distance  of  the  ends  18  feet? 


By  formula,   V=%7ra(R*+r*) 
-f-576)==9Xl476X7T=13284T==41732.9177626  cu.  ft, 


MENSURATION. 


285 


II 


III 


1.  30  ft.—  the  radius  of  the  larger  base. 

2.  24  ft.=the  radius  of  the  lesser  base. 

3.  18  ft.=the  altitude  of  the  frustum. 

4.  900  sq.  ft.=the  square  of  the  radius  of  the  lower  base. 

5.  576  sq.  ft.—  the  square  of  the  radius  of  the  upper  base. 

6.  1476  sq.  ft  =900  sq.ft.+576  sq.  ft.^their  sum. 

7.  .-.-1X18X^X1476=13284X^=41732.9177626  cu.  ft.= 

the  volume  ot  the  frustum  of  the  conoid. 
/.  The  volume  of  the  frustum  is  41732.9177626  cu.  ft. 
II.  THE  HYPERBOLIC    CONOID. 

1.   An  Hyperbolic  Conoid  is  the  solid  formed  by  revolv- 
ing an  hyperbola  about  its  axis  of  abscissa. 

Prob.  CXXIV.    To  find  the  surface  of   an  hyperbolic  con- 
oid, or  hyperboloid. 


Formula.—  S= 


^     ^ a  /» 

2*  O-J-          -4  dx=ln  /  - 
N  x2 — a2  J  a 


=TT~  }  x\e"x-—a2- 
a  ' 


a"~ 
e 


B-- 

a  ( 


*—  a2—  ab\- 


-log 


ab 


1( 

e 

Prob.  CXXV.    To  find  the  volume  of  an  hyperbolic  conoid. 

Formula. —  V=\n(R*-\-d'i  )h,  where  7?  is  the  radius  of 
the  base,  rtfthe  middle  diameter,  and  h  the  altitude. 

Rule. —  To  the  square  of  the  radius  of  the  base  add  the  square 
of  the  middle  diameter  between  the  base  and  the  vertex;     and  this 
sum  multiplied  by  the  altitude,  and  the  product  again  by  \7t, 
give  the  solidity. 

I.  In  the  hyperboloid  A  CB,  the  altitude 
CO  is  10,  the  radius  A  O  of  the  base  12, 
and  the  middle  diameter  DE  15.8745; 
what  is  the  volume? 

'l.'lO=the  altitude  CO. 

2.  12=the  radius  A  O  of  the  base. 

3.  15.8745=f  he  middle  diameter  DE. 

4.  144=122=the  square  of  the  radius 
II J  of  the  base. 

15.  251.99975=15. 87452=the  square  of  the  middle  diameter. 


FIG.  53- 


286  FINKEL'S  SOLUTION  BOOK. 

6.  395.99975=251.99975+144==the  sum   of  the   squares  of 

the  radius  of  the  base  and  the  middle  diameter, 

7.  .:.  £7TXlQX395.99975=2073.454691=the  volume. 
ill.      ,-.  The  volatile  of  the  conoid  is  2073.454G91. 

Prob.  CXXVI    To  find  the  volume    of  the  frustum  of  an 
hyperbolic  conoid. 


Formula.  —  ^=$?T0(7?2-f-^8+r*)»  where  7?  is  the  ra- 
dius of  the  larger  base,  and  r  tne  radius  of  the  lesser  base,  and  d 
the  middle  diameter  of  the  frustum. 

Rule.  —  Add  together  the  squares  of  the  greater  and  lesser 
semi  -diameters,  and  the  square  of  the  whole  diameter  in  the  mid- 
dle; then  this  sum  being  multiplied  by  the  altitude,  and  the  prod- 
uct again  by  -J-7T,  will  give  the  solidity. 


QUADI^ATUr^E  AND  CUBAXUr<E  OF 
SURFACES  AND  SOLIDS  OF  REVOLU- 
TION- 

1.  CYCLOID. 

Prob.  CXXVII.    To   find   the    surface  generated    by  the 
revolution  of  a  cycloid  about  its  base. 


,—S=2j27ryds=< 


\2r—y 
Rule.  —  Multiply  the  area  of  the  generating  circle  by  6^. 

Prob.    CXXVIII.    To  find  the  :  volume  of  the  solid  formed 
by  revolving'  the  cycloid  about  its  base. 


Formula.—  1= 


Rule.  —  Multiply  the  cube  of  the  radius  of  the  generating  cir- 
cle by  5?r2. 

Prob.    CXXIX.    To  find  the  surface    generated  by  revolv- 
ing" the  cycloid  about  its  axis. 


Formula.—  S=27ry^s=4:7ry=S7tr2  (  n—  f  ). 


Rule.  —  Multiply  eight  times  the  area  of  the  generating  circle 
by  n  minus  \. 

Prob.  CXXX.    To  find  the  volume  of  the  solid  formed  by 
revolving  the  cycloid  about  its  axis. 


MENSURATION.  287 

Formulu.—  V= 


Rule.  —  Multiply  \  of  the  volume  of  a  sphere   whose    radius  is 
that  of  the  generating  circle  by  f  /r2  —  f  . 

Prob.    CXXXI.    To  find  the  surface  formed  by  revolving 
the  cycloid  about  a  tangent  at  the  vertex. 

Let  Pbe  a  point  on  the  curve,  A£=P£=yJ  EP=A£=x, 
A  C=  CF=r,and  the  angle  A  CF 
=6.  Then  \ve  shall  have  AE= 
y  =  AC  —  C£=r  —  r  cos  6  ;  and 
A  B  =  x=FP+£F=  A  F-\-EF 
_r0_|_rsjn  B. 

.-.  Formula.  —  S= 


(r—  FIG.  54. 


=I§7tr*  J 


llllle.  —  Multiply  the  area  of  the  generating-  circle 

Prob.  CXXXI  1     To  find  the  volume  formed  by  revolving 
a  cycloid  about  a  tang-ent  at  the  vertex. 

Formula.—  V'^&Jny*  ctx=<27rC7r(r—rcosff)2r(l+ 

(l—  cos0~ 

=7r2r3=the     volume    generated    between    the 
curve  and  the  tangent. 


Rule.  —  Multiply  the  cube  of  the  radius  of  the  generating  cir- 
cle by  7zr2. 

2.  CISSOID.     . 

Prob.  CXXXIII.    To  find  the  volume  g-enerated  by  revolv- 
ing the  cissoid  about  the  axis  of  abscissa. 

Formula.  —  Y 


288 


FLNKEL'S  SOLUTION  BOOK. 


Prob.  CXXXIV.    To  find  the  volume  formed  by  revolving 
the  clssoid  about  its  asymptote. 

Formula.—  V=2jit(AR)*dy(Fig.  *j)=2*  (2«—  *)*X 


(  2iCt>  —  x  j  Y 

Prob.  CXXXV.    To  find  the  volume  formed  by  revolving 
the  Witch  of  Agnesi  about  its  asymptote. 


Formula. —  F—  / 7ty'2dx=\  ny2x — 


Prob.  CXXXVI.  To  find  the  volume  formed  by  revolving 
the  Conchoid  of  Nicomedes  about  its  asymptote,  or  axis  ol 
abscissa. 


Formula.—  V—  Cny^dx^^n  I  \  —  a_ 

J      L_          V£2  - 


2 


2.  SPINDLES. 

A  Circular  Spindle  is  the  solid  formed  by    revolving    the 
segment  of  a  circle  about  its  chord. 

Prob.  CXXXVII .    To  find  the  volume  of  a  circular  spindle. 

Let  AEBD  be  the  circular  spindle  formed 
by  revolving  the  segment  A  CBE  about  the 
chord  A  CB.  Let  AB=^a,  the  length  of 
the  spindle,  and  ED=^lb^  the  middle  diame- 
ter of  the  spindle.  Let  CI=K£=x,  the  ra- 
dius of  any  right  section  of  the  spindle,  and 

r=CZ,=y.     Then  the  required   volume    of 


the  spindle  is 


r 

=2rt 
J  o 


.  ..(1).  Let  R= 


-f-24..(2), 


be    the    radius    of   the    circle  and  9  the  angle  A  GB.     Then  by 
a  property  of  the    circle.    KI^^^R—  El]  y^EI,  or  y2 
El)  Y.EI.      But  JEI=EG—IG=R—(IC+CG)=R— 


whence  x=*fi2—y2  —  Rcosd  .  .  (3).     Substituting  this  value  of  A 
in  (1),  we  have   V= 


MENSURATION. 


289. 


Rule.  —  Multiply  the  area  of  the  generating  segment  by   tL 
path  of  its  center  of  gravity,  —  Guildin's  Rule. 

3.    THE   PARABOLIC  SPINDLE. 

A.  IPuTdbolic  Spindle  is  a  solid    formed  by  revolving  a 
parabola  about  a  double  ordinate  perpendicular  to  the  axis. 

Prob.    CXXXVIII.      To  find   the  volume  of  a  parabolic 
spindle. 

Formula.—  V=2  C  7t(h—xYdy=(l7r  f  (/fc2— 

Jo  Jo 


But  x= 


Rule. — Multply  the  volume  of  its  circumscribed  cylinder  by  r8^. 

I.  What  is  the  volume  of  a  parabolic  spindle  whose  length 
A  C  is  3  feet  and  height  BD  1  foot? 

By  formula,  F=if 7th*b=^7tX I2  X 3=4.9945484  cu.  ft. 
'1-  1  ft.— height     BD     of    the 
spindle. 

2.  3  ft=length  A  C. 

3.  ?rXl2X  3=9.42477795  cu.  ft. 

the  volume  of  its  circum- 
scribed cylinder. 
AX9.42477795  cu.   ft.= 
4.9945484  cu.  ft.,  the  vol- 
ume of  the  parabolic  spin- 
dle. FIG.  56. 
III.     .'.  The  volume  of  the  spindle  is  4.9945484  cu.  ft. 

Prob.  CXXXIX.  To  find  the  volume  generated  by  revolv- 
ing the  arc  of  a  parabola  about  the  tangent  at  its  vertex. 

Let  A  PC  be  an  arc  of  a  parabola  revolved  about  AB,  and  let 
P  be  any  point  of  the  curve.  Let  AE=PF=x,  and  AF=PF 
==yt  Then  the  area  of  the  circle  described  by  the  line  PF\&  nxz. 

.   Formula. —  V==. 


II. 


4. 


4^2  4/2 


290  FINKEL'S  SOLUTION  BOOK 

where  /i=the  height,  and  b=  CD,  the  or- 


dinate  of  the  curve. 

Rule. — Multiply  the  volume  of  its  circum- 
scribed cylinder  by  \. 

Prob.  CXLi.  To  find  the  volume  generated 
by  revolving  the  arc  APC  of  the  parabola 
about  BC  parallel  to  the  axis  AD. 

The  area  of  the  circle  generated  by  the  line 
GP  is  7t (b — -yY"> 

FIG   57. 

.-.  Formula. —  V=: 


Iillle.  —  Multiply  the  volume  of  its  circumscribed  cylinder  by  \. 

NOTE.  —  In  the  last  two  problems,  the  volume  considered,  lies  between 
the  curve  and  the  lines  AB  and  BC  respectively.  The  volume  generated 
by  the  segment  ACD  is  found  by  subtracting  the  volume  found  in  the  two 
problems  from  the  volume  of  the  circumscribed  cylinders. 

Prob.  CXLJ.  To  find  the  volume  formed  by  revolving  a 
semi-circle  about  a  tangent  parallel  to  its  diameter. 

Let  the  semi-circle  be  revolved  about  the  tangent  AG.       Let 
A  C=R  ,  PF=A  G=E  C=y,AF=  GP=x.    Then 
the  area  of  the  circle  generated  by  the  line  GP  is 

Ttx*.     But  *2=2,ff2—  ZR(RZ—  >'2)^—  y2;    for, 

;  whence 
and   X2=2fi2— 


Formula.—  V=2  JV 

-p—  _y2  )dy=$#  /?3  (  10—  STT),  which  is  the 

entire  volume  external  to  the  semi-circle- 

FIG.  59. 

Rule.  —  Multiply  one-fourth  of  the  volume  of  a  sphere  whose- 
radius  is  that  of  the  generating  semi-circle  by  (10  —  STT). 

XVI.     TEOULAI       SOLIDS. 


1.  A  Hef/ular   Solid  is  a  solid  contained  under  a  certain 
number  of  similar  and  equal  plane  figures. 

2.  The  Tetrahedron,  or  Regular  Pyramid,   is   a 

regular  solid  bounded  by  four  triangular  faces. 

3.  The  Hexahedron,  or  Cube,  is  a  regular  solid  bounded 
by  six  square  faces. 

4.  The   Octahedron  is  a  regular  solid   bounded   by  eight 
triangular  faces. 

5.  The   Dodecahedron    is  a  regular    solid    bounded   by 
twelve  pentagonal  faces. 


MENSURATION. 


291 


6.  The  Icosahedron  is  a  regular  solid  bounded  by  twenty 
equilateral  triangular  faces. 

These  are  the  only  regular  solids  that  can  possibly  be  formed. 

If  the  following  figures  are  made  of  pasteboard,  and  the  dotted 
lines  cut  half  through,  so  that  the  parts  may  be  turned  up  and 
glued  together,  they  will  represent  the  five  regular  solids. 


FIG.  59. 

1.    TETRAHEDRON. 
Prob.  CXLII.    To  find  the  surface  of  a  tetrahedron. 

Formula.  —  SW2V37where  /  is  the  length  of  a  linear 


side. 


—  Multiply  the  square  of  a    linear  side    by  ^3=1.7320 


508. 

I.     What  is  the  surface  of  a  tetrahedron  whose  linear  edge  is 
2  inches. 

By  formula,  S=/2V3=22V3=4V3==6.9282  sq.  in. 
-1.  2  in.=the  length  of  a  linear  side. 

side.  [surface. 

.=6.9282  sq.  in.,   the 
III.     .-.  The  surface  of  the  tetrahedron  is  6.9282  sq.  in. 

Prob.  CXLIII.    To  find  the  volume  of  a  tetrahedron. 


*  i 

{1.  2  in.=the  length  of  a  linear  side. 
2.  4  sq.  in.=22=the  square  of  a  linear  side 
3.  .-.  V?X  4  sq.  in.=1.73205x4  sq.  m.==6.l 


side. 


Formula.—  F=T\V2  /3,  where  /is  the  length  of  a  linear 


292  FINKEL'S  SOLUTION  BOOK. 

Rule.  —  Multiply  the  cube  of  a    linear  side  by  T^Vi,  or  .11785. 

I.     Required  the  solidity  of  a   tetrahedron    whose    linear   side 
is  6  feet? 

By  formula,  V=^2  /3==TVV2x63=18V2=25.455843  cu.  ft. 


.  6  ft.=the  length  of  a  linear  side. 

II.  <  2.  216  cu.  ft.^the  cube  of  the  linear  side. 

3.  .-.  T12V2"x216  cu.  ft.=V2xl8  cu.  ft.=25.45843  cu.  ft. 

III.  .-.  The  volume  of  the  tetrahedron  is  25.45843  cu.  ft. 

2.  OCTAHEDRON. 

Prob.  CXLJV.    To  find  the  surface  of  an  octahedron. 
Formula.—  5=2^3  /2  . 


5. — Multiply  the  square  of  a  linear  side  by  2*J~3 ,  i.  e.,    by 
two  times  the  square  root  of  three. 

I.     What  is  the  surface  of  an    octahedron   whose  linear  side  is 
4  feet? 

By  formula,  S=2*/3  /2=2V3X/2=32V3=55.4256  cu.  ft. 

1.  4  ft.=the  length  of  a  linear  side. 

2.  16  sq.  ft.:=42— the'  square  of  the  linear  side. 

'<°>.  /.  2V3X16  sq.   ft.=V3x32  sq.  ft.=1.7320oX32  sq.  ft.= 

55.4256  sq.  ft. 
III.     .-.  The  surface  of  the  octahedron  is  55.4256  sq.  ft. 

Prob.  CLXV.    To  find  the  volume  of  an  octahedron. 

Formula. —  F=^V2  /3 

Rule. — Multiply  the  cube  of  a  linear  side  by  JV#,  i.  e.,  by  one- 
third  of  the  square  root  of  two. 

I.      What  is  the  volume  of  an   octahedron  whose  linear  side  is 
8  inches? 

By  formula,   V=#fa  /3^V2x83==.4714045X512==241.359104 
cu.  in. 

1.   g  in.— che  length,  of  a  linear  side. 
].  512  cu.  in.— 83=the  cube  of  a  linear  side. 
5.  .-.iV/'2x512cu.  in  =^X  1-4142135X512  cu.  in.= 
241. 359104  cu.  in. 

III.   .  .  The  volume  of  the  octahedron  is  241.359104  cu.  iu. 

3.  DODECAHEDRON. 
Prob.  CXLVI.     To  find  the  surface  of  a  dodecahedron. 


MENSURATION.  293 


Formula.—  5= 


Rule.  —  Multiply  the   square  of  a    linear  side  by 
,   or  20.64.57285. 


I.     What  is  the  surface  of  a  dodecahedron  whose  linear  side  is 
3  feet? 


By  formula,  6-= 

185.8115565  sq.  ft. 

.  3  ft.=the  length  of  a  linear  side. 
2.  9  sq.  ft.=32=square  of  a  linear  side. 


IL3.  .-.  15x9  sq.  ft=20.6457285x9  sq.  ft. 


=185.8115565  sq.  ft. 
III.     The  surface  of  the  dodecahedron  is  185.8115565  sq.  ft. 

Prob.  CXLVII.    To  find  the  volume  of  a  dodecahedron. 


Formula.—  F= 


Rule 


.  —  Multiply  the  cube  of  a  linear  side  by  5+\{  -  —  jjg  --  V 

or  7.663115. 

I.     The   linear   side  of  a  dodecahedron  is  2  feet  ;    what  is  its 
volume  ? 


By  formula,   K= 
=61.20492  cu.  ft. 

II.  2  ft.=the  length  of  a  linear  side. 
2.  8  cu.  ft.=22=cube  of  a  linear  side. 
3.  .-.  5V[A(47+2lV5)]x8cu.  ft.=7.663115x8cu.  ft. 
=61.20492  cu.  ft.,  the  volume. 

III.     .-.  The  volume  of  the  dodecahedron  is  61.20492  cu.  ft. 

4.     ICOSAHEDRON. 

Prob.  CXI,  VIII.    To  find  the  surface  of  an  icosahedron. 
Formula.—  5=5Vs/2  =8.66025  X  /2  . 


Rule.  —  Multiply    the    square    of    a    linear   side  by   5V^,  or 
3.66025. 


294 


FINKEL'S  SOLUTION  BOOK. 


I.  What  is  the  surface  of  an  icosahedron  whose  linear  side 
is  5  feet. 

By  formula,  5=5\/3/2=5V3x52=125V3=216.50625  sq.  ft. 

(1.  5  ft.=length  of  a  linear  side, 
jj  J2.  25  sq.  ft.=52=the  square  of  a  linear  side. 

]  3.  .'.  5^3X25  sq.  ft.=8.66025x25  sq,  ft.=216.50625  sq.  ft. 

=the  surface. 
III.     /.  The  surface  of  the  icosahedron  is  216.5062^  sq.  ft. 

Prob.  CXLIX.    To  find  the  solidity  of  an  icosahedron. 
Formula.—  ^=jV[i(  7+3V§ )]  /3=2. 18169  X  /3 . 

'Rule.— Multiply  the  cube  of  a  linear  side  by  fV[|(7+#VS)], 
or  2.18169 

I.  What  is  the  volume  of  an  icosahedron  whose  linear  side  i& 
3  feet? 

By  formula,  F=|V[i(7+3V5)]/3=2.18169  X33=58.90563  cu.ft. 

II.  3  ft.— the  length  of  a  linear  side. 
2.  27  cu.  ft.=:33— the  cube  of  a  linear  side. 
3-  .'.  |V[i(7+3V5)]x27cu.  ft.=2. 18169 X 27  cu.  ft. 
=58.90563  cu.  ft.=the  volume. 

III.     .-.  The  volume  of  the  icosahedron  is  58.905G3  cu.  ft. 
NOTE. — The  surface  and  volume  of  any  of  the  five  regular  sol- 
ids may  be  found  as  follows  : 

Rule  ( 1 ). — Multiply  the  tabular  area  by  the  square  of  a  linear 
side,  and  the  product  will  be  the  surface 

Rule  (2). — Multiply  the  tabular  volume  by  the  cube  of  a 
linear  side,  and  the  product  will  be  the  volume. 

Surfaces  and  volumes   of  the   regular  solids,  the  edge  being  1.. 


NO.  OF 
SIDES. 

NAMES. 

SURFACES. 

VOLUMES. 

4 

Tetrahedron 

1.73205 

0.11785 

6 

Hexahedron 

6.00000 

100000 

8 

Octahedron 

3.46410 

0.47140 

12 

Dodecahedron 

20.64578 

7.66312 

20 

Icosahedron 

8.66025 

2.18169 

XVII.   PRISMATO1D. 

1.  A.  IPrismatoid  is  a  polyhedron  whose  bases  are  any  two 
polygons  in  parallel  planes,  and  whose  lateral  faces  are  triangles 
determined  by  so  joining  the  vertices  of  these  bases,  that  each 
lateral  edge,  with  the  preceding,  forms  a  triangle  with  one  side 
of  either  base 


MENSURATION.  295 

2.  A  Prismoid  is  a  prisrnatoid  whose  bases  have  the  same 
number  of  sides,  and  every  corresponding  pair  parallel. 

Prob.  CL».    To  find  the  volume  of  any  prism  at  old. 

Formula  («).—  ^=i«(^1+3^4|a)=ia(^2+3^/2a),  where 
a  is  the  altitude,  B  ^  the  area  of  the  lower  base,  A^a  the  area  of  a 
section  distant  from  the  lower  base  two-thirds  the  altitude,  J3Z 
the  area  of  the  upper  base,  and  A\a  the  area  of  a  section  distant 
two-thirds  the  altitude  from  the  upper  base. 

Remark.  —  This  simplest  Prismoidal  Formula  is  due  to  Prof. 
George  B.  Halsted,  A.  M.,  Ph.  D.,  Professor  of  Mathematics  in 
the  University  of  Texas,  Austin,  Texas,  who  was  the  first  to 
demonstrate  this  important  truth.  The  formula  universally  ap- 
plies to  all  prisms  and  cylinders;  also  to  all  solids  uniformly 
twisted,  e.  g.  the  square  screw;  also  to  the  paraboloid,  the  right 
circular  cone,  the  frustum  of  a  paraboloid,  the  hyperboloid  of  one 
nappe,  the  sphere,  prolate  spheroid,  oblate  spheroid,  frustum  of 
a  right  cone,  or  of  a  sphere,  spheroid,  or  the  elliptic  paraboloid, 
the  groin,  hyperboloid,  or  their  frustums.  For  a  complete 
demonstration  of  the  Prismoidal  Formula,  see  Halsted  's  Elements 
of  Geometry  or  Halsted'  s  Mensuration. 

Rule.  —  (a)  Multiply  one-  fourth  its  altitude  by  the  sum  of  one 
base  and  three  times  a  section  distant  from  that  base  two-thirds 
the  altitude. 

Formula  (6).  V=\a(B^M^-B^,  where  a  is  the  alti- 
tude, BI  and  J32  the  areas  of  the  lower  and  upper  bases  respect- 
ively, and  Af  the  area  of  a  section  midway  between  the  two 
bases. 


mid 


Rule.  —  (£)   Add  the  area  of  the  tivo   bases   and  four  times  the 
d  cross-section;  multiply  this  sum  by  one-sixth  the  altitude. 


XVIII.     CYLINDRIC    ICINGS. 


1.  A  Cylindric  Hing  is  a  solid  generated  by  a  circle 
lying  wholly  on  the  same  side  of  a  line  in 
its  own  plane  and  revolving  abont  that  line. 
Thus,  if  a  circle  whose  center  is  O  be  re- 
volved about  DC  as  an  axis,  it  will  gener- 
ate a  cylindric  ring  whose  diameter  is  AB 
and  inner  diameter  2  BC.  OC  will  be  the 
radius  of  the  path  of  the  center  O. 

FIG.  60. 
Prob.  CLJ.    To  find  the  area  of  the  surface  of  a  solid  ring1. 

Formula.  —  6'=27rrX27r7?=47r2r7?,  whare  r  is  the  ra- 
dius of  the  ring,  and  R  is  the  distance  from  the  center  of  the  ring 
to  the  center  of  the  inclosed  space. 


296  FINKEL'S  SOLUTION  BOOK. 

Rule. — Multiply  the  generating  circumference  by  the  path  of 
its  center.  Or,  to  the  thickness  of  the  ring  add  the  inner  diame- 
ter and  this  sum  being  multiplied  by  the  thickness,  and  the  pro- 
duct again  by  9.8697044  will  give  the  area  of  the  surface. 

I.     What  is  the  area  of  the  surface  of  a  ring  whose  diameter 
is  3  inches  and    the  inner   diameter  12 
inches. 

By  formula,  6'=4^V/?=47r2  Xl|X 
(Iff  6)=*r 2  X 45=9.8696044X45 
—444.132198  sq,  in. 

1.  l-£in.=-J-  of  3  in.=the  radius  r 

of  the  ring. 

2.  6  in.=^  of  12  in.=the  radius  of 

the  inclosed  space. 

3.  6  in.-|-l-J-  in.=7-J-  in.  =  the    ra- 
il J  dius  R  of  the  center  of  the  FIG,  61. 

ring. 

4.  nA  C=7r3=the  circumference  of  a  section. 

5.  7r7Jr==27T/<9=27r7|=7T  15=the  path  of  the  center. 

6.  .-.  7r3Xtfl5=7r  245=444.132198  sq.  in.=the  area  of  the 

surface  of  the  ring. 
III.     .-.  The  area  of  the  surface  of  the  ring  is  444.132198  sq.  in. 

Prob.  CLJI.   To  find  the  volume  of  a  cylindric  ring. 

Formula. —  £=7r  V2/?=?rr2  y^nR,  where  r  is  the  ra- 
dius AI  of  the  ring,  and  R  the  distance  from  the  center  of  the 
ring  to  the  center  of  the  inclosed  space. 

Rule. — Multiply  the  area  of  the  generating  circle  by  the  path 
of  its  center.  Or,  to  the  thickness  of  the  ring  add  the  inner  di- 
ameter, and  this  sum  being  multiplied  by  the  square  of  half  the 
thickness,  and  the  product  again  by  9. 8696044 »  wifl  give  the 
volume. 

I.  What  is  the  volume  of  an  anchorring  whose  inner  diame- 
ter is  8  inches,  and  thickness  in  metal  3  inches? 

By  formula,    V=7r*r*R=7t*  X(H)2  X(3+8)=24.75X 
9.8696044=244.2727089  cu.  in. 

1.  1-|  in.=J  of  3  in.=the  radius  of  the  ring. 

2.  8  in.=the  inner  diameter. 

3.  4  in.-|-l|-  in.=5|-  in.=the  radius  R   of   the    path    of  its 
,-  i  center. 

4.  7r(l-|-)2=the  area  of  the  generating  circle. 

5.  27r(5|)=7T  xll=the  path  of  its  center. 

6.  .-.  7rllX7r(H)2=7r2X 24.75=9.86044X24.75 

=244.2727089  cu.  in.,  the  volume  of  the  ring, 
'III,     /.  The  volume  of  the  ring  is  244.2727089  cu.  in. 


MENSURATION.  297 

THEOREM  OF  PAPPUS. 

If  a  plane  curve  lies  wholly  on  one  side  of  a  line  in  its  own 
plane,  and  revolving  about  that  line  as  an  axis,  it  generates 
thereby  a  surface  of  revolution,  the  area  of  which  is  equal  to  the 
product  of  the  length  of  the  revolving  line  into  the  path  of  its 
center  of  mass  ;  and  a  solid  the  volume  of  which  is  equal  to  the 
revolving  area  into  the  length  of  the  path  described  by  its  center 
of  mass. 

XIX.     MISCELLANEOUS     MEASURE- 
MENTS. 

1.  MASONS'  AND  BRICKLAYERS'  WORK. 

Masons'  WOrfc  is  sometimes  measured  by  the  cubic  footr 
and  sometimes  by  the  perch.  A  perch  is  16-^  ft.  long,  1-J-  ft.  wide, 
1  ft.  deep,  and  contains  16£xHXl=24f  cu.  ft. 

Prob.  CLJII.  To  find  the  number  of  perch  in  a  piece  of 
masonry. 

Rule. — Find  the  solidity  of  the  wall  in  cubic  feet  by  the  rules 
given  for  the  mensuration  of  solids,  and  divide  the  product  by  2^. 

I.  What  is  the  cost  of  laying  a  wall  20  feet  long,  7  ft  9  in. 
high,  and  2  feet  thick,  at  75  cts.  a  perch. 

1.  20  ft.=the  length  of  the  wall, 

2.  7  ft.  9  in.=7|  ft.=the  height  of  the  wall,  and 

3.  2  ft.=the  thickness. 


II. 


4.   .-.  20X71X2—310  cu.  ft.=the  solidity  of  the  wall. 


5.  24f  cu.  ft=l  perch. 

6.  310  cu.  ft.=310-5-24i==12H  perches. 

7.  75  cts.=the  cost  of  laying  1  perch. 

.'.  12ff  X75cts.=$9.39if=the  cost  of  laying  12ff  perches. 
III.     .-.   It  will   cost   $9.39^1  to  Iayl2ff  perches   at  75  cts.  a 
perch. 

2.  GUAGING. 

Gauging  is  finding  the  contents  of  a  vessel,  in  bushels, 
gallons,  or  barrels. 

Prob.  CLJV.    To  gauge  any  vessel. 

Rule. — Find  its  solidity  in  cubic  feet  by  rules  already  given; 
this  multiplied  by  1728-^-215042  or  .83,  will  give  the  contents  in 
bushels;  by  1728-^-231.  will  give  it  in  wine  gallons,  which  divided 
by  Sl\  will  give  the  contents  in  barrels. 

Prob.  CL.V.  To  find  the  contents  in  gallons  of  a  cask  or 
barrel. 

Rule. — (1)  When  the  staves  are  straight  from  the  bung  to 
each  end;  consider  the  cask  two  equal  frustums  of  equal  cones^ 
and  find  its  contents  by  the  rule  of  Proh.  XCIII. 


298  FINKEL'S  SOLUTION  BOOK 

(2).  When  the  staves  are  curved;  Add  to  the  head  diameter 
{inside}  two-tenths  of  the  difference  between  the  head  and  bung 
diameter;  but  if  the  staves  are  only  slightly  curved,  add  six- 
tenths  of  this  difference;  this  gives  tiie  mean  diameter;  express 
it  in  inches,  square  it,  multiply  it  by  the  length  in  inches,  and  this 
product  by  .0084  ;  the  product  will  be  the  contents  in  wine  gallons. 

3.  LUMBER  MEASURE. 

Prob.  CL.VI.    To  find  the  amount  of    square-edged  inch 
boards  that  can  be  sawed  from  a  round  log. 


Doyle's  J^tile.  —  From  the  diameter  in  inches  subtract 
four;  the  square  of  the  remainder  will  be  the  number  of  square 
feet  of  inch  boards  yielded  by  a  log  16  feet  long. 

I.  How  much  square-edged  inch  lumber  can  be  cut  from  a 
log  32  in.  in  diameter,  and  12  feet  long? 

1.  32  in.=the  diameter  of  the  log. 

2.  12  ft.=the  length. 

3.  32  in.  —  4  in.=28  in.=the  diameter  less  4. 


II. 


4.  844  ft.=282=the  square  of  the    diameter  less  4,    which 


by  the  rule,  is  the  number  of  feet  in  a  log  16  ft.  long. 

5.  12  ft.=f  of  16  ft. 

6.  .'.  |  of  844  ft.=633  ft=the  number    of   feet    of   square- 

edged  inch  lumber  that  can  be  cut  from  the  log. 

III.  .'.  The  number  of  square-edged  inch  lumber  that  can  be 
cut  from  a  round  log  32  inches  in  diameter  and  12  ft.  long  is 
633  ft. 

4.     GRAIN  AND  HAY. 

Prob.  CI/VII.  To  find  the  quantity  of  grain  in  a  wagon 
bed  or  in  a  bin. 

Rule.— Multiply  the  contents  in  cubic  feet  by  1728-^-2150.42, 
or  .83. 

I.  How  many  bushels  of  shelled  corn  in  a  bin  40  feet  long, 
16  feet  wide  and  10  feet  high  ? 

1.  40ft.=the  length  of  the  bin. 

2.  16  ft.=the  width  of  the  bin,  and 


II. 


3.   10  ft.=the  height  of  the  bin. 


4.  .'.  40X16X10—6400  cu.  ft.=the  contents  of  the  bin  in 

cu.  ft.. 

15.  .-.  6400  X -83  bu.=5312  bu.=the  contents  of  the  bin  in  bu. 
III.     .-.  The  bin  will  hold  5312  bu.  of  shelled  corn. 

Rule. — (1)  For  torn  on  the  cob,  deduct  one-half  for  cob. 
(2)   For  corn  not    "shucked"   deduct   two-thirds  for  cob    and 
shuck. 


II. 


I.  How  many  bushels  of  corn  on  the  cob  will  a  wagon  bed 
hold  that  is  10J  feet  long,  3^  feet  wide,  and  2  feet  deep? 

1.  10i  ft.=the  length  of  the  wagon  bed,' 

2.  3£  ft.=its  width,  and 

3.  2"ft.=its  depth.  [in  cu.  ft 

4.  .'.  10iX3|X2=73^cu.   ft.=contents  of  the  wagon  bed 

5.  ,'.  73^  X-8  bu=58.8  bu.=n umber  of  bushels  of  shelled 

corn  the  bed  will  hold. 

6.  .'.   £  of   58.8   bu=29.4  bu.=the    number  of  bushels    of 

corn  on  the  cob  that  it  will  hold. 

III.     /.  The  wagon  bed   will  hold  29.4  bu.  of  corn  on  the  cob. 

Prob.  CLVIII.  To  find  the  quantity  of  hay  in  a  stack,rick, 
or  mow. 

Rule. — Divide  the  cubical  contents  in  feet  by  550  for  clover  or 
by  4^0 for  timothy;  the  quotient  will  be  the  number  of  tons. 

Prob.  CLXIX.    To  find  the  volume  of  any  irreg-ular  solid. 

Rule. — Immerse  the  solid  in  a  vessel  of  water  and  determine 
the  quantity  of  water  displaced. 

I  A  being  curious  to  know  the  solid  contents  of  a  brush 
pile,  put  the  brush  into  a  vat  16  feet  long,  10  feet  wide,  and 
8  feet  deep  and  containing  5  feet  of  water.  He  found,  after 
putting  in  the  brush,  that  the  water  rose  1-J-  feet ;  what  was  the 
contents  of  the  brush  pile? 

1.   16  ft.=the  length  of  the  vat, 
TT  J2.   10  ft.=the  width,  and 

><!3.   1|  ft.=the  depth  to  which  the  water  rose. 

4.   .'•'.  16X10X11=240  cu.  ft.=the  volume  of  the  brush  pile. 
III.      .'.   240  cu.  ft=the  volume  of  the  brush  pile. 

XX.     SOLUTIONS    OK  MISCELLANEOUS 
PROBLEMS. 

Prob.  CLX.  To  find  at  what  distance  from  either  end,  a 
trapezoid  must  be  cut  in  two  to  have  equal  areas,  the  divid- 
ing- line  being1  parallel  to  the  parallel  sides. 


Formula.—  ^=^^ 

where   A   is  the  area  of  the  trapezoid,  b  the 


lower   base,    and  bl  ,  the  upper  base.  \  ^(bz  -\-b\)   is   the  length  of 
the  dividing  line. 

Rule.  —  1-  Extract  the  square  root  of  half  the  sum  of  the 
squares  of  the  parallel  sides  and  the  result  will  be  the  length  of 
the  dividing  line. 


300  FINKEL'S  SOLUTION  BOOK. 

2.  Divide  half  the  area  of  the  whole  trapezoid  by  half  the  sum- 
of  the  dividing  line  and  either  end,  and  the  quotient  will  be  the' 
distance  of  the  dividing  line  from  that  end. 

I.      I  have   an   inch  board  5  feet   long,  17    inches   vvidc%  at  one 
end  and  7  inches  at  the  other;  how  iar  from  the 
large  end  must  it  be  cut  straight  across  so  that 
the  two  parts  shall  be  equal? 

By  formula,  </=-£•  i 


==^(17_|_7)60-;-[V^(172+72)+17]=720-:-30 
=24  in.=2  ft. 

(  1.  Let  AB CD  be  the  board,  [end, 

2.  AB=ll  in.=fl,  the  width  of  the  large 

3.  Z>C=7  in.=y,  the  width  of  the  small 

end,  and  [board. 

4.  HK=5  ft.=60  in.==0,  the  length  of  the 

5.  Produce  HK,  AD,  and   BC  till   they    _ ___ 

meeting.   Then  by  similar  triangles,    *'       plG   $2. 

8.  IE  GL=&ED  C+AB  CD—ED  C+ED  C+AB  C  D 


II. 


9.   .'.  EGL=%(EDC+EAB),  i.  e.,  EGL  is  an  arithme- 
tic mean  between  EAB  and  EDC. 

10.  .-.   G!Z2=:^(^4^2+Z>C2)=|(^2+^/2)—  an  arithmetic 

mean  between  EAB  and  EDC, 

11.  G:Z=' 


12.  Draw  CM  perpendicular  to  AB. 


13.  EL=  %GL= 

14.  IL=FL—FI 

15.  CM=HK=a. 


16.  MB^^b—'b'}.       Then  in  the   similar  triangles    CMB 

and  CIL,  _ 

17.  MB\IL\\CM\CI,  01-  |(^-^ 

C7.     Whence 

18.  C7=  « 


60^      v'     'J    '        '=36  in. 


17—7 

=3  ft. 
19.   .-.  7M=CM—Cf=5  ft. — 3  ft.==2  ft,,,  the  distance  from 

the   large   end  at  which  the  board  must  be  cut  in  two 
IS       to  have  equal  areas. 

III.     .-.  The  board  must  be  cut  in  two,,  at  a  distance  of  2  feet 
from  the  large  end,  to  have  equal  areas  in  both  parts. 

(R.  H^  A»^f.  4P7,  greb.  101.), 


MENSURATION. 


301 


Prob.  CLiXI.    To  divide  a  trapezoid  into  n  equal  parts  ami 
find  the  leiig'th  of  each  part. 


Formula.—  &!== 


the  width  of  the  small  end,  b  the  width  of  the  large  end,  and  a 
the  length  of  the  trapezoid.  hl  is  the  length  of  the  first  part  at 
the  small  end,  hz  the  length  of  the  second  part,  and  so  on. 

I.  Aboard  AB  CD  whose  length  BC  is  36  inches,  width 
AB  8  inches  and  DC  4  inches,  is  divided  into  three  equal  pieces. 
Find  the  length  of  each  piece. 

(n—  l) 


a 
By  formula,  £1=—  — 

—  1)42+82—  4]=9[V32—  4]=36(V2—  1)  =14.911686  in. 


:36[2— 

1 


II. 


V3]=9.6462  in. 
4in.=the    width    Z>C  of   the  small 

end, 
8  in.=the  width  ^4^  of  the  large  end, 

and 

36  in.=the  length  BC  of  the  board. 
.;.  216  sq.  m.=%(A£+DC)x£C 

=i(8+4)  X36=the  area  of  the 

board. 
^  of  216  sq.  in.=72  sq.  in.=the  area  of 

each  piece. 
AK=AB—KB(=DC)=%  in.—  4  in. 

=4  in.       In    the    similar    triangles 

AKD  and  DCE,  FIG  63, 

AK\DK\\AB\BE,  or  4  in.:36  in.::8  \n.\BE.     Whence, 
^^=(36x8)H-4=72in.  [triangle  ABE. 


ABE—ABCD=2$&    sq.  in.— 216  sq-   in.=72   sq.  in. 
=area  of  the  triangle  DCE. 


302  FINKEL'S  SOLUTION  BOOK. 

11.  DCE+DCGF^Z   sq.   in.+72    sq.   in.  =  144  sq.  in. 

=the  area  of  the  triangle  FGE. 

12.  DEC+DCGF-\-FGIH=^  sq.   in.+72  sq.  in.+72 

sq.  in.=216  sq.  in.—  the  area  of  the  triangle  HIE. 

13.  FEG\DEC\\EG*\EC*,<x 

144  sq.  in.:72sq.  in.::  G^2:362.     Whence, 

14.  G-fi'=^(144x362)-*-72==36V2=50.911686  inches. 

15.  /.  GC=GE—  C^=50.911686  in.—  36  in.=14.911686 

in.  ,  the  length  of  FG  CD.     Again, 

16.  DEC:HIE\:EC*:El*,or 

72  sq.  in.:216sq.  in.::362  :^/2.     Whence, 

17.  .57==V(216x362)-r-72==  36X^3=62.3538  in._ 

18.  .-.  GI=EI—  ^£=36^3—  36V2=36(A/3—  \/2) 

=11.442114  in.,  the  length  ofHIGF,  and 

19.  £I=EB=EI=Tb—  36V3=36(2-V3)=9.6462  in.,  the 

length  tfABIH. 
^7=9.6462  in., 
III.  ,-J  G/=11.442114  in.,  and 
lGC=14.911686m. 

Prob.  CLXII.    To  find  the  edge  of  the  largest  cube  that 
c  an  be  cut  from  a  sphere. 


^^j—  ==iV  3Z>=.57735  X  A    where  D 
is  the  diameter  of  the  sphere. 

Rule.  —  Divide  the  square  of  the  diameter  of  the  sphere  by 
three  and  extract  the  square  root  of  the  quotient;  or,  multiply  the 
diameter  by  .57785. 

I.  What  is  the  edge  of  the  largest  cube  that  can  be  cut  from  a 
sphere  6  inches  in  diameter? 


By  formula,  e=^—  =1y=6x       =%  X  6^.57735X6 

=3.4641  in. 

{1.  6  in.=the  diameter  of  the  sphere. 
2.  .'.  .57735X6  in.=3.4641  in.=the  edge  of  the  largest  cube 
that  can  be  cut  from  the  sphere. 

III.     .'.  The  edge  of  the  largest  cube    that  can    be  cut    from  a 
sphere  whose  diameter  is  6  inches,  is  3.4641  in. 

Prob.  CLXIII.    To  find  the  edge  of  the  largest  cube  that 
can  be  cut  from  a  hemisphere. 


Formula.—  *=-=\6X.£>—  408248  X^>. 


*  —  Divide  the  square  qf  the  diameter  by  6,  and  extract  the 
square  root  of  the  quotient  ;  or,  multiply  the  diameter  by  .Jf.08248* 


MENSURATION.  303 

I.     What  is  the  edge  of  the  largest  cube  that  can  be  cut  from  a 
hemisphere,  the  diameter  of  whose  base  is  12  inches? 


By    formula,  e=V Z>2-^-6=V^=12V  %=$V  6Xl2=.408248 
X  12=4.899176  in. 

yj  f  1.  12  in.— the  diameter  of  the  base  of  the  hemisphere. 
|E.  .-.  .408248  X 12  in.=4.899176  in. 

III.  .-.  The  edge  of  the  largest  cube  that  can  be  cut  from  a 
hemisphere,  the  diameter  of  whose  base  is  12  feet,  is  4. 899176  in. 

Prob.  CL.XIV.  To  find  the  diameter  or  radius  of  the  three 
largest  equal  circles  that  can  be  inscribed  in  a  circle  of  a 
given  diameter  or  radius. 

Formula.— d=D+(  1+f  A/3  )=Z>^-2.1557=  .4641  X# 

Kule. — Divide  the  diameter  or  radius  of  the  given  cirele  by 
2.1557  and  the  quotient  will  be  the  diameter  or  radius  of  the  three 
largest  equal  circles  inscribed  in  it;  or,  multiply  the  diameter  or 
radius  by  .464!,  and  the  result  will  be  the  diameter  or  radius  re- 
spectively of  the  required  circles. 

I.  A  circular  lot  15  rods  in  diameter  is  to  have  three  circular 
grass  beds  just  touching  each  other  and  the  larger  boundary  ; 
what  must  be  the  distance  between  their  centers,  and  how  much 
ground  is  left  in  the  triangular  space  about  the  center? 

By  formula,  2r=2^?-7-(l+|V/3)==2^?-r-2.1557=^.T13Vr 
=6.9615242  rd.=the  distance  between  their  centers. 

Construction. — Let  AHE  be  the  circular  lot,  C  the  center,  and 
A  CE  any  diameter.  With  E  as  a  center  and  radius  equal  to 
CE  describe  an  arc  intersecting  the  circumference  of  the  lot  in  H. 
Draw  a  tangent  to  the  lot  at  E  and  produce  the  radius  CH  to 
intersect  the  tangent  at  B.  Bisect  the  angle  CBE  and  draw 
the  bisector  GB.  It  will  meet  the  radius  CE  in  G,  the  center 
of  one  of  the  grass  beds.  Draw  GF  perpendicular  to  CB.  Then 
GF=GE,  the  radius  of  one  of  the  grass  beds.  Draw  EH. 
Then  EH=CH~EC,  and  CH=HB,  because  the  triangle 
EHB  is  isosceles. 

1.    CZ£=7-|  rd.==#,  the  radius  of  the  lot. 

3.  EB^VCBI—CE^VCIR^—R^RVZ.    In  the 

similar  triangles  CFG  and  CBE, 

4.  CF\FG\\CE\\EB,w   CF:GF:\R\RV%.     But 

5.  CF= 

^6.  .'.  7?(2— VZ)\GF\:R'M^%-     Whence, 


.4641=3.48075  rd.=the  radius. 


304 


FINKEL'S  SOLUTION  BOOK. 


11. 


=2r=2X(23— 3)=6.9615rd.,  the  distance  be- 
tween their  centers. 


2. 


3. 


area    of  the  triangle 
IGK. 
Area  DKF=\  of  the 

small  circle,  because 
the    angle'   DKF    is 
60°,  or  \  of  360°. 
±.  .'.  Area 

5.  -J7rr2:= 

=the    area     of    the  FIG.  64 

three  parts  of  the  small  circles  within  the  triangle 

IGK. 

6.  .-.  r2A/3—  i7rr2=^2(V/3—  |7r)=.16125368r2 

—  .16125368X[^(2V3—  3)]2=.16125368X('21 

—  12A/3)./?2=:.16125368x.2153904x^2 
=.03473265    X^2=-03473265x(7^)2  =1.953712 
sq.  rd.=the  area  of  the  space  inclosed. 

{6.9615  rd.=the  distance  between  their  centers,  and 
1.953712  sq.  rd.=the  area  inclosed  about  the   center  of 
the  given  lot.  (R.  H.A.,  p.  4.07,  prob.  100.) 

Prob.  CL.XV.  Having-  given  the  area  inclosed  by  three 
equal  circles  to  find  the  radius  of  a  circle  that  will  just  in- 
close the  three  equal  circles- 


~Nv  03473265>)' 


where  A  is  the  area  inclosed- 


Rvile.  —  Divide  the  area  inclosed  "by  .034.73265  and  extract  the 
square  root  of  the  quotient,  and  the  result  will  be  the  radius  of  the. 
required  circle. 

Prob.  CLXVI.  Having  given  the  radius  a,  b,  c,  of  the 
three  circles  tangent  to  each  other,  to  find  the  radius  of  a 
circle  tangent  to  the  three  circles. 

abc 

Formula.  —  r  or  r'=n          .    .  —  r-j  —  •:—:-  -  j-r^r> 

1\/[abc(a+b+c)]  =F  (  ab-\-ac-\-bc) 

the  minus  sign  giving  the  radius  of  a  tangent  circle  circumscrib- 
ing the  three  given  circles  and  the  plus  sign  giving  the  radius 
of  a  tangent  circle  inclosed  by  the  three  given  circles. 

NOTE.  —  This  formula  is  due  to  Prof.  E.  B.  Seitz,  Late  Professor  of  Mathe- 
matics in  the  North  Missouri  State  Normal  School,  Kirksville  ,  Mo.,  of 
whom  we  give  a  biographical  sketch  accompanied  by  his  photograph. 

This  for  mula  is  taken  from  the  School  Visitor,    Vol.  II.  p.  Ill,   with  the 


MENSURATION.  305 

slight  change  that  the  plus  sign  is  introduced  for  the  case  in  which  the 
tangent  circle  is  inclosed  by  the  three  given  circles.  The  problem  of  finding 
two  circles  tangent  to  three  mutually  tangent  circles,  is  one  supposed 
to  have  been  proposed  by  Archimedes  more  than  2000  years  ago,  though 
the  problem  he  proposed  was  not  so  general  —  the  diameter  of  one  of  the 
given  circles  being  equal  to  the  sum  of  the  diameters  of  the  other  two. 

The  problem  of  finding  all  circles  that  can  be  drawn  within  three  mu- 
tually tangent  circles  and  tangent  to  each  of  them,  has  been  simply  and 
elegantly  solved  by  D.  H.  Davison,  Minonk,  111.  The  above  formula  led 
him  to  his  wonderful  solution.  For  a  complete  and  elegant  solution, 
where  he  has  actually  computed  and  constructed  81  circles  tangent  to  three 
given  circles,  see  School  Visitor,  Vol.  VI  ,  p.  80. 

Prob.  CLXVII.  To  find  the  surface  common  to  two  equal 
circular  cylinders  whose  axes  intersect  at  right  angles. 

Formula.  —  S—16jR2,  where  R  is  the  radius  of  the  cylinders. 

Rule.  —  Multiply  the  square  of  the  radius  of  the  intersecting" 
cylinders  by  J6. 

I.  If  the  radius  of  two   equal  circular   cylinders,  intersecting 
at  right  angles  is  4  feet,  what  is  the  surface  common  to  both? 

By  formula,  S=  16^?2  =  16  X42  =256  sq.  ft. 

f  1.  4  ft.  =  the  radius  of  the  cylinders. 

II.  J  2.   16  sq.  ft.  =  42  =  the  square  of  the  radius  of  the  cylinders 
)  3.  .-.  16  X  16  sq.  ft.  =  256  sq.  ft.  =  the  surface  common  to  the 

two  cylinders. 

III.  .•.  256  sq.ft.  =  the  surface  common  to  the  two  cylinders. 

Prob.  CLXVIII.  To  find  the  volume  common  to  two  equal 
circular  cylinders  whose  axes  intersect  at  right  angles. 


Formula.  —  V=  V^8»  wnere  R  *s  tne  radius  of  the 
cylinder. 

Rule.  —  Multiply  the  cube  of  th0  radius  of  the  cylinders 
ly  V. 

1.  A  man  digging  a  well  3  feet  in  diameter,  came  to  a  log  3 
feet  in  diameter  lying  directly  across  the  entire  well;  what  was 
the  volume  of  the  part  of  the  log  removed  ? 

By  formula,    V=  \fR*  =  V(l)8  =  18  cu-  ft 

1.  3  ft.  =  the  diameter  of  the  log  and  the  well. 

2.  11  ft.  =  the  radius. 


II 


3.  3| cu.  ft.  =  ( 1 1) 3  =  the  cube  of  the  radius. 


4.  .•.  V6  X  3f  cu.  ft.=  IScu.ft. ,  the  volume  of  the  part   of 

the  log  removed. 
III.  .'.  The  volume  of  the  part  of  the  log  removed  is  18  cu.ft. 

Prob.  CLXIX.  To  find  the  height  of  an  object  on  the 
earth's  surface  by  knowing  its  distance,  the  top  of  the  ob- 
ject being  visible  above  the  horizon. 


306  FINKEL'S  SOLUTION  BOOK. 

Let  J3F=a  be  any  object,  AJ3=t  a  tangent  to  the  earth's  sur- 
face from  the  top  of  the  object,  and  FE-=D  the  diameter  of  the 
earth.  Then  by  Geometry,  AB*=BF(BF+FE),m  t'1=a(a 

t* 
-{-/)).     .*.  a=     t   ^-.     But  a  is  very  small  as  compared  with    the 

diameter  of  the  earth  and  AB=AF  without  appreciable  error. 

AFZ     c* 
.'.   Formula.  —  a—  —  =  —  =-rr,  where  c  is   the  distance  to 

the  object  from  the  point  of  observation. 

I2 
When  c=\  mile,  a=          =-|  ft.,  nearly. 


Rule.  —  Multiply  the  square  of  the  distance  in 
miles  by\,  and  the  result  'will  be  the  height  of  the 
object  in  feet  • 

I.  What  is  the  height  of  a  steeple  whose  top  can 
be  seen  at  a  distance  of  10  miles? 

c2       102        102 
By  formula,  «=== 


{1.  10  miles=the  distance  to  the  steeple. 
2.  100=102=the  square  of  the  distance. 
3.  .-.  f  of  100=  66|  ft.=the  height  of  th 


III.  .-.  The  height  of  the  steeple  is  66f  ft. 

Prob.  CLXX.  To  find  the  distance  to  an  object  by  kno\v- 
ing  its  height,  the  top  only  of  the  object  being  visible  above 
the  horizon. 


Rule.  —  Multiply  the  height  of  the  object  in  feet  ~by  |  and  ex- 
tract the  square  root  of  the  product  ,  and  the  result  'will  be  the  dis- 
tance in  miles. 

I.  At  what  distance  at  sea  can  Mt.  Aconcagua  be  seen,  if  its- 
height  is  known  to  be  24000  feet? 

By  formula,  c=Vj^=V%  X24000=V36000=190  mi.,   nearly. 

il.  24000  ft.=the  height  of  the  mountain 
2.  1X24000=36000. 
3.  .-.  V  36000=10  V360=190  mi.,  nearly. 
III.   /.  Mt.  Aconcagua  can  be  seen  at  a  distance  of  190    miles. 

Prob.  CXXXI.  Given  the  sum  of  the  hypotenuse  and 
perpendicular,  and  the  base,  to  find  the  perpendicular. 

S2  _  ^2 

Formula.  —  /==  -  -  —  ,   where  s   is   the   sum    of  the  hy- 
potenuse and  perpendicular,  and  b  the  base. 


MENSURATION.  307 

Rule. — 1-  From  the  square  of  the  sum  of  the  hypotenuse 
and  perpendicular  subtract  the  square  of  the  base,  and  divide  the 
difference  by  twice  the  sum  of  the  hypotenuse  and  perpendicular. 

2.  To  find  the  hypotcmise:  To  the  square  of  the  sum  of  the 
hypotenuse  and  perpendicular,  add  the  square  of  the  base  and  di- 
vide this  sum  by  twice  the  sum  of  the  hypotenuse  and  perpendic- 
ular. 

I.  A  tree  120  feet  high  is  broken  off  but  not  severed.  The 
top  strikes  the  ground  34  feet  from  the  foot  of  the  tree;  what  is 
the  height  of  the  stump? 

52_£2      1202— 342  [stump. 

By  formula,  /=      ^        :___=55J£  ft.,  the  height  of  the 

1.  120  ft.=the  sum  of  the  hypotenuse  and  perpendicular. 

2.  34  ft.=the  base,  or  the  distance  the  top  strikes  from  the 

foot  of  th~  tree. 


II. 


3.   14400  sq.  ft.=1202=the  square  of  said  sum, 


4.  1156  sq.  ft.=342=the  square  of  the  base,  and 

5.  14400  sq.  ft—  1156  sq.  ft.=13244  sq.  ft.=the  difference. 

6.  .-.  13244-r-(2Xl20)=55^  ft.=the  height  of  the  stump. 

III.     .-.  The  height  of  the  stump  is  55^  feet. 

NOTE.  —  This   rule  is  easily  derived  from  an  algebraic   solution.      Thus: 
Let  #=the  perpendicular,  s  —  #—  the  hypotenuse,  and  fcrthe  base.     Then, 


x*-\-b*=(s—  *)2,  or  xz-\-l>2=s*—  2sx+x*,  and  y—    ~. 

Prob.  CLXXII.  To  find  at  what  <listance  from  the  large 
end  of  the  frustum  of  a  right  pyramid,a  plane  must  he  passed 
parallel  to  the  hase  so  that  the  two  parts  shall  have 
equal  solidities. 

3  V 
Formula.  —  h=-  -  -  ,     where    V  is   the 


volume  of  the  frustum,  B  the  area  of  the  lower  base,  B2  the  area 
of  the  "dividing  base,"  andVj2J52  the  area  of  the  mean  base  be- 
tween the  "dividing  base"  and  and  lower  base. 

Rule.  —  1.     Find  the  -volume  of  the  frustum  by  Prob.  XCIII. 

2.  find  the  dimensions  of  the  "dividing  base"  by  extracting  the 
cube  root  of  half  the  sum  of  the  cubes  of  the  homologous  dimensions 
of  the  upper  and  lower  bases.      Then  find  the  area  of  the  ^divid- 
ing base" 

3.  Divide  half  the  volume  of  the  frustum  by  one-third  of  the 
sum  of  the  areas  of  the  lower  base,  "dividing  base,"  and  mean 
base  between    them,   and  the  quotient  will  be  the  length  of  the 
lower  part. 

I.  How  far  from  the  large  end  must  a  stick  of  timber,  20  feet 
long,  5  inches  square  at  one  end  and  10  inches  square  at  the 
other,  be  sawed  in  two  parts,  to  have  equal  solidities? 


FINKEL'S  SOLUTION  BOOK. 


By  formula,  /i=—j — - 


240(102+10x5+52) 


42000 


1680 


2(100+25^36+^2-^6)  8+2^36+^6 
1680  1680 


8+6.603855+5.4513618     20.0552168 


7e     3     . 


h$  piece  of  limber,  ABCl> 
the  lower  base,  EFGH  the  upper  base,  and  OL  the  altitude. 
Prolong  the  edges  AH,  BE,  CF,  and  DG  and  the  altitude  <9Z 
till  they  meet  in  P.  Draw  KL  to  the  middle  point  of  AD,  Ol 
to  the  middle  point  of  GH  and  draw  PIK.  Let  SMNR  be 
the  dividing  base. 

1.  ^4^=10  iti.=£,    the  side  of   the   lower 

base. 

2.  HE=Z>    in.=<:,  the    side  of    the   upper 

base,  and 

3.  OZ=20  ft.=240  in.=«,  the  altitude. 

4.  KQ=KL—  QL(=IO)=\(b—  c)=i(10 

in.  —  5  in)=2-Jin.   By  similar  triangles, 

5.  ^^:  QI::KL:PL,  or 

Whence, 

ft. 


II. 


/. 

8. 
9. 

10. 
11. 
12. 


b  —  <: 


—  c 


=20  ft. 


FIG   66 
=2000  cu.  in.,  the  volume  of  the  pyramid  HEFG. 


13. 
14. 


+c2)=14000    cu.   in.,    the    volume  of   the    frustum 

ABCD—E. 
.-.  £F=£of  14000  cu.  in.=7000cu.   in.,  the    volume  of 

each  part. 
V+-J  F==2000   cu,   in.+7000  cu.  in.=9000  cu.  in.,   the 

volume  of  the  pyramid,  SMNR  —  P,  and 
v+  K=2000  cu.  in.+14000   cu.    in.=16000  cu.  in.,  the 

volume  of  the   pyramid  ABCD  —  P.     By  the  princi- 

ple of  similar  solids,  \AB*,  or 

HEFG-P  :  SMNP-P  :  ABCD-P  :  :  HE*  :  SM*  : 
v  :  v         V-.V      V.  :  c3  :  5J/3  :  b*  .      But 


MENSURATION  309 

15.  ,4-1  V==fiv+(v+  F)],  i.  e.,  *-H  V,  or  SMNRP\s  an 
arithmetical    mean  between  v  and  *;-J-  Fj   or  HEPG 


17.  .-.  SM*=±(c?  +&*),{.  e,  $/J/3  is  an  arithmetical  mean 

between  ./7£3  and  ABr>,  or  c3  and  £3.      Whence, 

18.  5^=== 

8.2548188-Mn. 

19.  SJ/2=y[|  (^-f  £3)  ]  2=(  fy  36  )2=Y'  ^6=68.  14202 

sq.  inr=the  area  of  the  dividing  base. 
20  V(S^X^^2)=SJ/x^#==if/36xlO=25y36== 

82.  54818  sq.    in.=the    area  of   the    mean  base  of  the 

part  cut  from  the  frustum. 
21.  /. 


+§y  (36)  XlO+(fy36)2]=i£:T(100-f  82.54818 
+68.14202)=  ^ZT^X  250.6902  =-Zrx  83.5634  =  the 
volume  of  the  fru&tumA£CD—M    .But  [—  M. 

23.  |  F=7000  cu.  in.==the  volume  of   the  frustum  ABCD 

24.  .'.  LTX 83.5634=7000  cu.  in.     Whence, 

25.  Zr==7000-^83.5634==83.76883  in.==6  ft.    11.76883  in., 

the  length. 

III.  .'.  The  stick  must  be  cut  in  two  at  a  distance  of  83.76883 
:in.,  or  6  ft.  11.76883  in.,  from  the  large  end. 

NOTE. — The  frustum  of  a  cone  may  be  divided  into  two  equal  parts  in 
the  same  manner.  The  frustum  of  a  pyramid  or  a  cone  can  be  divided  into 
any  number  of  equal  parts  on  the  same  principle  as  that  for  dividing  a 
trapezoid  into  n  equal  parts,  Prob.  CLXI. 

I.  The  area  of  a  rectangle  whose  length  is  20  rods  is  120  sq. 
rods;  what  is  the  area  of  a  similar  rectangle  whose  length  is  30 
xods? 


>, — Similar  areas  are  to  each  other  as  the  squares 
•of  their  like  diinentions  or  as  the  squares  of  any  other  homologous 
dines. 

'1.   20  rods=the  length  of  the  given  rectangle,  and 

2.  120  sq.  rd.=its  area. 

3.  30  rods=the  length  of  the  required  rectangle. 
J4.   .'.  202:302  ::120sq.  rd.  :  (?).    Whence, 

[5.    ?= (120x302)-r-202=270  sq.  rd. 
•III.  .•.  The  area  of  the  rectangle  is  270  sq.  rd. 

I.     The  area  of  a  rectangle  whose  width  is  7  feet,  is  210  sq.  ft.  ; 
what  is  the  length  of  a  similar  rectangle  whose  area  is  2100  sq.  ft. 

1.  210  sq.  ft  =the  area  of  given  rectangle,  and 

2.  7  ft.=its  width.     Then 
TT    3.  210-7-7=30  ft —its  length. 

'    4.  /.  210  sq.  ft:  2100  sq.  ft.  :  :  302  :(  ?).      Whence, 
5.    ?=(2100x302)-:-210=300    ft.=the   length    of   the    re- 
quired  rectangle. 


II, 


310  FINKEL'S  SOLUTION  BOOK 

III.      The  length  of  the  required  rectangle  is  300  feet. 

I.  If  the  weight  of  a  well  proportioned  man,  5  feet  in  height, 
be  125  Ibs.,  what  will  be  the  weight  of  a  similarly  proportioned 
man  6  feet  high? 

Principle. — Similar  solids  are  to  each  other  as  the  cubes  of 
their  like  dimensions  or  as  the  tubes  of  any  other  homologous  lines. 

1.  5  ft.=the  height  of  the  first  man,  and 

2.  125  lbs.=his  weight. 

3.  6  ft— the  height  of  the  second  man. 

4.  /.   53  :63  ::1251bs:(?).      Wivnce, 

5.  ?=(125X63H-53=216  Ibs.,  the  weight    of   the    second 

man. 
III.      .-.The  weight  of  the  man  whose  height  is  6  feet,  is  216  Ibs. 

I.  James  Page  has  a  circular  garden  10  rods  in  diameter.  How 
many  trees  can  be  set  in  it  so  that  no  two  shall  be  within  10  feet 
of  each  other  and  no  tree  within  2-J-  feet  of  the  fence? 

Construction. — LetAJB  C  be  the  circular  garden,  AC  it  diameter, 
and  O  its,  center.  Then  with  O  as  a  center  and  radius  A  O  =  -J-of 
(10X16|- ft— 2X2|  ft.),  or  80  ft,  discribe  the  circle  abcdef, 
and  in  it  describe  the  regular  hexagon  abcdef.  Then  aO 
=tf&=80  ft.  Begin  at  the  center  of  the  circle  and  put 
8  trees  10ft.  apart  on  each  radii,  aO,  60,  cO,  dO.  eO,  and/0. 
Then  joining  these  points  by  lines  drawn  parallel  to  the  diame- 
ter of  the  circle  as  shown  in  the 
figure,  their  points  of  intersec- 
tion will  mark  the  position  of  the 
trees.  Hence,  the  trees  are  ar- 
ranged in  hexagonal  form  about 
the  center.  The  first  hexagonal 
row  contains  6  trees,  the  second, 
12,  the  third  18,  and  so  on.  Since 
the  radius  of  the  circle  on  which 
the  trees  are  placed  is  80  feet  and 
the  trees  10  feet  apart,  there  will 

be  8  hexagonal  rows. 

FIG.   67, 

1.  6=the  number  of  trees  in  the  first  hexagonal  row. 

2.  12=the  number  of  trees  in  the  second  hexagonal  row. 

3.  48=the  number  of  trees  in  the  eighth  hexagonal  row. 

4.  .•.  216=4(6+48)X8=the  number  of  trees   in  the  eight 


II, 


hexagonal  rows. 

5.  24=6x4=the  number  of  trees  at  the  sides  of  the  hexa- 

gon abcdef. 

6.  .'.  216+24+1,  the  tree  at  the  center ,=241=the  number 

of  trees  that  can  be  set  in  the  garden. 


MENSURATION. 


311 


III.     /.  There  can  be  set  in  the  garden,  241  trees. 

(  Greenleafs  Nafl  Arith.,  />.  444,  prob.  25.} 

I.  There  is  a  ball  12  feet  in  diameter  on  top  of  a  pole  60  feet 
high.  On  the  ball  stands  a  man  whose  eye  is  6  feet  above  the 
ball;  how  much  ground  beneath  the  ball  is  invisible  to  him? 

Construction.—  Let  BE  be  the  pole,  L  the  center  of  ball,  and 
A  the  position  of  the  man's  eye.  Draw  AFC  tangent  to  the  ball 
atT?  and  draw  LFwdBC.  Then  the  trbnglc  AFL  is  right- 
angled  at  F. 

'1.  60  it.=BE,  the   length  of   the 
pole. 

2.  12  ft.=J?Z>,   the    diameter     of 

the  ball,  and 

3.  6  ft.=A£>,  the  height  of  the 

man's  eye  above  the  ball. 
12    ft.—  AD+DL=AL.    Now 


II. 


6. 


7. 


8. 


=V(122  —  62)=6\/3ft.  In  the 

similar    triangles    ALF  and 
ACB, 
AF-.LF:  :AB:   BC,  or 

6V3  ft.  :6ft.  ::(6ft.+12ft. 

+60  ft.),  or  78  ft.:  BC. 
.-.  BC=(  6x78)-5-6V3=78 

-2-V3=jx78V3=26V3  ft. 


the  circle  over  which  the  man  can   not  see. 


FIG  68. 

sq.  ft.=the  area  of 


III.  .-.  6371.1498932  sq.  ft.=the  area  of  the  invisible  ground 
beneath  the  ball. 

I.  Three  women  own  a  ball  of  yarn  4  inches  .in  diameter. 
How  much  of  the  diameter  of  the  ball  must  each  wind  off,  so  that 
the  may  share  equally? 

1.  4  in.=the  diameter  of  the  ball.     Then 

2.  i-7r(4)3==\27r=the  volume  of  the  ball. 

3.  ij-  of  3327r=3g27T=each  woman's  share. 

4.  3-^7T — 3927r=6g47T=the  volume  of  the  ball  after  the  first 

has   unwound  her  share.  But 

±7fjD'd==the  volume  of  any  sphere  whose  diameter  is  />. 
i6^47r.    Whence, 


II. 


6. 

8.  Z>=^/9l|8^4^|=l4^18=4X2.6207414=3.4943219  in., 

diameter  of  the  ball  after   the   first  unwound  her  share. 

9.  .'.  4  in. — 3.4943219  in=.5056781  in.,  what  the  diameter 

was  reduced  by  the  first  woman. 

30.  -££-7t—*g7t=*-g7t,  the  volume  of  the   ball     after     the 
second  had  unwound  her  share. 


-312  FINKEL'S  SOLUTION  BOOK. 


11.  .-.  y(V*r-H?r)=4yi=fy  9=JX2.0800837 

=2.5734448  in.,  the  diameter  of  the  ball  after  the  sec- 
ond woman  unwoud  her  share. 

12.  .-.  3.4943219  in.—  2.5734448  in.=.7208771  in.,  what  *he 

diameter  was  reduced  by  the  second  woman. 

The  diameter  was  diminished    .5056781  in.  by  the  first 

woman, 

.7208771  in.  by  the  second  woman,  and 
2.7734448  in.  by  the  third  woman. 

(Milne's  Prac.  Arith.,p.  335,  prob.  8.) 

NOTE.  —  The   following   are  the  formulas  to  divide  a  sphere   into  n  equal 
parts,  the  parts  being  concentric: 


*nd 


is  the  diameter  of  the  sphere;  Z^the  diameter  after  the  first 
part  is  taken  off;  Z>2,  the  diameter  after  the  second  part  is  taken 
off;  and  soon.  Then/? — Z>1?  D^ — Z)2,  &c,  are  portions  of 
the  diameter  taken  off  by  each  part. 

I.     A  park  20  rods  square  is  surrounded  by  a  drive  which  con- 
tains j1^-  of  the   whole   park;   what  is  the 
width  of  the  drive? 

1.  20rd.=^4/>=DC,  a  side  of  the 

park. 

2.  400  sq.  rd.=202= the  area  of  the 

park  A  BCD. 

3.  TVo-  of  400  sq.  rd.=76  sq.  rd.=the 

-II. ' 


area  of  the  path. 


4.  400  sq.  rd.— 76  sq.  rd«=324  sq.  rd. 

=the  area  of  the  square  EFGH.  FIG.  69. 

5.  ^^=V(324)=18  rd.,  the  side  of  the  square  EFGH. 

6.  .'.  7/T— .^^=20rd.— 18rd.=2rd.,  twice  the  width  of  the 

path. 

17.  .'.  1  rd.=|  of  2  rd.=the  width  of  the  path. 
III.     .-.  The  width  of  the  path  is  1  rod. 

I.  My  lot  contains  135  sq.  rd.,  and  the  breadth  is  to  the  length 
as  3  to  5  ;  what  is  the  width  of  a  road  which  shall  extend  from 
one  corner  half  around  the  lot  and  occupy  \  of  the  ground. 

Construction.— Let  ABCD\>z  the  lot,  and  DABSNR  the 
road.  Produce  AB,  till  BE  is  equal  to  AD.  Then  AE  is 
•equal  toAB-\-AD.  OnA£,  construct  the  square  AEFG,  and 


MENSURATION. 


313, 


on  EF  and  GF respectively,  lay  off  El  and  T^VTequal  to  AB. 
Then  construct  the  rectangles  BEIH,  ILKF,  and  KMDG. 
They  will  each  be  equal  to  A  BCD,  for  their  lengths  and  widths 
are  equal  to  the  length  and  width  of  ABCD.  Continue  the  road 
around  the  square.  Then  the  area  of  the  road  around  the  square  is 
four  times  the  area  of  the  road  DABSNR. 

1.  £=the  width  AD  of  the 

lot.     Then 

2.  f=the  length  AB. 

3.  1x1=135  Sq.   rd.,    the 

area  of  the  lot. 

4.  £Xt=T    of  135    sq.  rd. 

=27  sq.  rd.,  and 
5-  fXf=(f)2= 3  times 27 
sq.rd.=81  sq.  rd. 

6.  /.  i=V81= 9rd.,the  « 

width  AD, 

7.  £=£  of  9  rd.=3  rd.,  and 

8.  4=5  times  3  rd.=15 

rd.,the  length  AB. 

9.  15  rd+9  rd.=24  rd.=  FIG.  70. 

AjB,the  side  of  the  square  AEFG. 

10.  .-.  576  sq.  rd.=242=  the  area  of  the  square   AEFG. 

11.  33|  sq.  rd.=i  of  135  sq.  rd.=the  area  of  the  road 
DABSNR. 

12.  .-.   135  sq.  rd.=4x33j  sq.   rd.=the     area    of   the   road 
around  the  square.     Then 

13.  576  sq.  rd. — 135  sq.  rd.=441  sq.  rd.,    the    area    of   the 

square  NOPQ. 

14.  .-.  21rd.=V441  =NO,  a  side  of  the  square  NOPQ. 

15.  AE — NO=24;  rd. — 21  rd.==3rd.=twice    the  width  of 

the  road. 

16.  .'.   H  rd.=24|  ft.=-J-  of  3  rd.=the  width  of  the  road. 

III.     /.  The  width  of  the  road  is  24|  ft. 

(jR.  H.  A.,p.407,pr0b.99.) 

I.  The  length  and  breadth  of  a  ceiling  are  as  6  to  5  ;  if  each 
dimension  were  one  foot  longer,  the  area  would  be  304  sq.  ft. ; 
what  are  the  dimensions? 

Construction. — Let  ABCD  be  the  ceiling,  AB  its  width  and 
BC  its  length.  Let  AIGE\>t  the  ceiling  when  each  dimension 
is  increased  one  foot.  On  BC,  lay  ofF^^fequal  to  AB  and  draw 
L,K  parallel  to  AB.  Then  ABKL  is  a  square  whose  side  is. 
the  width  of  the  ceiling. 


314 


FINKEL'S  SOLUTION  BOOK 


II. 


1.  l=AB,  the  width  of  the  ceil- 

ing.  'Then 

2.  \=BC,  the  length,  and 

3.  -fxf=^^X^C=the  area  of 

the  ceiling. 

4.  fXl=^CX#/,    El  being  1 

foot,=the  area    of   the    rect- 
angle B  CHI. 

5.  |Xl= DCxCF,   CF  being 

1  foot,=the  area  of  the  rect- 
angle DCFE. 

6.  1  sq.ft.=!2=the    area  of  the 
square  CFGH. 

7-   .-.fXf+fXl+fXl+lsq.ft. 
=the  area  of  AIGE.      But 


9.  . 

10. 
11. 
12. 
13. 


DCFE, 

f+f ,  <>r 

fxf+fxi+-5 


14. 
15. 

16. 

17. 

18. 

19. 


20. 
21. 

22. 


23. 
24. 


i.e.,  it   equals    a    rectangle    whose   length  is 
TT1 ,  and  width  1  ft. 

IXl+lsq.  ft.=fXf+y  Xl+1  sq.  ft. 
=the  area  of  AIGE.     But 
304  sq.  ft.=the  area  of  AIGE. 
.'•  f  Xf+y  Xl+l  sq.  ft.=304sq.  ft. 
I  Xf+y  X  1=303  sq.  ft=the  area    of 
y=y  Xf »  in  which  y  is  y  ft.  ;  for  a  rectangle  whose 
length  is  y ,  and  the  width  1  ft,  has   the  same  area    as 
a  rectangle  whose  width  is  y  ft.  and  length  1,  orf. 
•'•  f  Xf +y  Xf=303  sq.  ft,  in  which  y  is  y  ft 

ixf+Hxf=50i  sq.  ft.=j.  of  (fxf+yxf)=i  of 

303  sq.  ft., 


Whence, 
AIHCFE.Kut 


sq.  ft.     But 
|xf=the  area  of  the  square  ABKL,  and 
y  Xf—  the  area  of  the  rectangle  ALNP  whose  length 

AL  is  f  and  width  LN  y  ft. 
iof(¥Xt)=HXf=haIf  the  rectangle  ALNP=the 

rectangle   OMNP,    which   put  to   the  side  AB  of  the 

square  ABKL  as  in  the  figure. 

|Xf+y  Xf=252^  sq.  ft.=thearea  of  SRAOMK. 
\\\  sq.  ft.=(J-J)2==theareaof  the  square  RQOA,  since 

AR  is  \\  ft. 
...fxf+yXf+Hi  sq.    ft.=252i  sq.    ft.+ifi  sq.   ft. 

=^648i  sq    ft—the  area   of  (SRAOMK+RQOA). 

=the  area  of  SQMK. 

f+H  ft.=V-flli=:1T¥  ft.=the  side   5"^  of  the  square 

SQMK. 
iHVV  ft.—  Ji-  ft.=V2°  ft.=15  ft.= 

=Afit  the  width  of  the  ceiling. 


MENSURATION. 


315 


TTT 


25.  i=4  of  1-5  ft.= 3  ft.,  and 

^26.  f=6  times  3  ft.=18  ft.=  7?C,  the  length  of  the    ceiling. 
C  15  ft.=the  width  of  the  ceiling,  and 
\  18  ft.=the  length. 
Remark. — In  this  solution  there  is  but  one  algebraic  operation  ; 
viz.,    extracting    the    square  root   of  the   trinomial    expression, 
(|X|+y  Xf+yfi  sq-  in-)>  in  steP  23.     This    might  have  been 
omitted   and  then  the  solution  would  have  been  purely  arithmeti- 
cal ;  for,  the  area  of  the  square  SQMK  being  known,    as  shown 
by  step  22,  its  side  SK  could  have  been  found  by  simply  extract- 
ing the  square  root  of  its  area,2-^!-1-  scl-  ^-     Then   by  subtracting 
SB,  which  is  \\  ft,  from  SK,  we   would    get   BK(=AB),    the 
width  of  the  ceiling. 

The  following  solution  is  quite  often   given  in  the  schoolroom: 
304  -=-(5X6)  =  10+.      >/10  =  3+. 
5  X3  = 15,  the  width  and  6  X  3=18,  the  length. 

I.  A  tin  vessel,  having  a  circular  mouth  9  inches  in  diameter, 
a  bottom  4-j  inches  in  diameter,  and  a  depth  of  10  inches,  is  £  part 
full  of  water  ;  what  is  the  diameter  of  a  ball  which  can  be  put 
in  and  just  be  covered  by  the  water? 

Construction. — Let  AB  CD  be  a  vertical  section  of  the  vessel, 
AB  the  top  diameter,  DC  the  bottom  diameter,  and  EF  the  al- 
titude. Produce  AD,  BC,  and  EF\\\\  they  meet  in  G.  Draw 
MC  parallel  to  EF.  In  the  triangle  A  CB  inscribe  the  largest 
circle  IEP  arid  let  Q  be  its  center.  Draw  the  radius  IQ.  Now 
(  1.  AE=^AB=R=^  in.=the  radius  of  the  mouth. 

2.  CF=±DC=r=2±  in.,  the  radius    of   the  bottom,  and 

3.  EF=a=lO  in.,  the  altitude  of  the  vessel. 


—  2£  in.=2J  in.  In  the  similar  triangles 


5.  MB-.MC-.  :EB\EG,or 

R—r  la-.iR-.EG.     Whence, 


*=20   in.,  the  altitude 


of  the  triangle  A  GB. 

1&AGB 
*     v~ 


AB+A  G+BG~AB+A  G+BG 


4JRa 


==3f  in.,  the   radius 
of  the  largest  sphere  that  can  be  put  in  the  vessel  or  in 


316  FINKEL'S  SOLUTION  BOOK. 

the  cone  A  GB. 

10. 

=the  volume  of  the  largest    sphere    that   can  be   put: 
in  the  cone  AGB. 

11.  £  EGxx££2=i7t2aR2=$7f  X20X(4|)2=1357r,    the 
volume  of  the  cone  A  GB. 


12.  ...  i*^-|*  ^.^  X 


the  quantity  of  water  in  the  cone  which  will  just  cover 
the  largest  ball  that  can  be  put  in  the  cone  A  GB. 

13.  ^7r^GX^C2=i^r2=^7rXlOX(2i)2=-i-f^7r,     the 

volume  of  the  cone  DGC. 

14.  .-.  i7Ttf7-2+i  of  the  volume  of  the  vessel  =  if^+i   of 

the  volume  of  the   vessel—  the    quantity  of   water    in 
the  cone  necessary  to  cover  the  required  ball.     But 

15.  $7ta(R*+Rr+r*  )=i7rlO[(4|)2+4|X2i+(2i)2] 

£f  5.7T,  the  volume  of  the  vessel,  by  Prob.  XCIII. 

16.  .*.     nar2--    °f  tne  volume  of  the  vessel  =rtar2--±  of 


7T-|~|-of  -9-4r§-7T=^-|f£-7T,  the  quantity  necessary  to  cover 
the  required  ball. 

17.  .'.  The  quantity  of  water  necessary  to  cover  the  largest 
ball:  the  quantity  of  water  necessary  to  cover  the 
required  ball  :  :  (radius)3  of  largest  ball  :  (radius)3  of 
required  ball.  Hence, 


19.  9_o^9^  :i4|5^  .  :  (3|)3  :JfO*.     Whence, 

20.  y337:|^55^3    :HO.     Whence, 


21.  JYO= 


and 

22.  18f  (^)=6.1967-hin.,  the   diameter    of  the  required 
ball. 

III.     /.  The  diameter  of  the  required  ball  is  6.1967+  in. 

I.  I  have  a  garden  in  the  form  of  an  equilateral  triangle 
whose  sides  are  200  feet.  At  each  corner  stands  a  tower;  the 
height  of  the  first  tower  is  30  feet,  the  second  40  feet,  and  the 
third  50  feet.  At  what  distance  from  the  base  of  each  tower 


MENSURATION 


317 


must  a  ladder  be  placed,  so  that  without  moving  it  at  the  base  it 
may  just  reach  the  top  of  each,  and  what  is  the  length  of  the 
ladder? 

Construction. — Let  ABC  be  the  triangular  garden  and  AD, 
BE,  and  CF  the  towers  at  the  corners.  Connect  the  tops  of  the 
towers  by  the  lines  ED  and  DF. 
From  G  and  H,  the  middle  points 
of  DB  and  DF,  draw  GJ/and  HN 
perpendicular  to  DE  and  DF,  and 
at  M  and  N  draw  perpendiculars  to 
AB  and  A  C  in  the  triangle  ABC, 
meeting  at  O.  1  lien  O  is  equally 
distant  from  D  and  E.  For,  since 
M  is  equally  distant  from  D  and  E, 
and  MO  perpendicular  to  the  plane 
ABED,  every  point  of  MO  is 
equally  -distant  frc  i  D  and  E.  For 
a  like  reason,  every  point  of  NO  is  p[G 

equally  distant  frcl_i  D  and  F\  hence,  O  their  point  of  intersec- 
tion, is  equally  dJ^ant  from  D,  E,  and  F  and  is,  therefore,  the 
point  where  the  ladder  must  be  placed.  Draw  _Z?/and  DJ  par- 
allel to  AB  and  A  C,  G^f  and  HL  perpendicular  to  AB  and  A  C, 
MP  perpendicular  to  A  C,  and  OR  parallel  to  NP.  Draw  the 
lines  OB,  OC,  and  OA,  the  required  distances  from  the  base  of 
the  ladder  to  the  bases  of  the  towers.  Draw  EO,  the  length  of 
the  ladder. 


1.  AB=£  C=A  C=200  ft.=j,  the  side  of  the  triangle. 

2.  ^C=50  ft.=0,  the  height  of  the  first  tower, 

3.  -Z£/?=40  ft.=$,  the  height  of  the  second  tower,  and 

4.  AD=Zb  ft.=c,  the  height  of  the  third  tower.    Let 

5.  h=*J[AB*—(±A  C)2]=V[>2—  (i*)2]=  iV3*=100 

Y/3  ft=the  perpendicular  from  J?  to  the  side  A  C. 

6.  &f=J3£—£f=AD=6—c=40ft.—3Qft. 


=10  ft. 

:jB+AD)=\(b+c)=$(4Q  ft.+30  ft.) 
=35  ft.  In  the  similar  triangles  DIE  and  GKM, 
DI-.IE  :  :  GK\  KM,  or  5  :  b— c  :  :  $(b+c) :  KM. 

*—  c2_4Q2—  302          f 

~Zs 2X200""  ~~  f     '' 


7.   G^= 


8. 


10.  AM= 

= 101  j  ft,  and 


=98^  ft.     In  like  manner, 
12.  J7£=i(«+c)=i(50  ft.+30  ft.)=40  ft., 


318 


FINKEL'S  SOLUTION  BOOK. 


II. 


,2 r2 


13.  LN=     ^    —4  ft, 


14.  AJV= 


15.  JVC=AC—AN=s 


2s  2s 

+        o  o  Of 

gg— ga  j2-f( 

2s  < 


=104  ft. 


96  ft.       By  similar  triangles, 
' 


16. 

:  AP.       Whence, 
17.  AP= 
18-  .-. 


19.  J?O=PN= 

—C2  )^.2s 
similar  triangles, 

20.  AB'.BL:  :AM:MP,ors: 

MP.     Whence, 

21.  AfP= 
22. 


[lar  triangles, 
Bysimi- 


23.  XM=(s*+2a2—  ^2 

c2  )-f-125]V3=17HV3  ft.     Again 

24.  MP-.MA  '.:RO-.  OM, 


or 


:  OM. 


25.  .-.  OyT/^ 

ft 

26.  O  N=R  P=MP—R  M= 


2  2  ^V  2 


-f-65]v/3=33|  V3  ft-     Then 

27.  OC 

(9 
L 

=111.8796+ft. 

28.  0^= 


=V14116TL=1  18.81  11+ft. 


29. 


B.  =^ 

length  of  the  ladder. 


MENSURATION. 


319 


1.  111.8796-fft.=the  distance  from  base  of  the  ladder- 

to  the  base  of  the  tower  FC, 

2.  1 18.81 11+ft.— the  distance  from  the  base  of  the  lad- 
Ill,  .'.'i         <Jer  to  the  base  of  the  tower  AD. 

'3.  115.8278+ft.=the  distance  from  the  base  of  the  lad- 
der to  the  base  of  the  tower  BE,  and 
4.   122.5402+ft.=the  length  of  the  ladder. 

(  Greenleaf's  Nat* I  Arith.,  p.  444,  prob.  38.) 
Remark. — When  the   sides  of  the  triangle  are  unequal,  pro- 
ceed in  the  same  manner  as  above.       In  some    cases    the    base  of 
the  ladder  will  fail  without  the  triangle. 

I.  At  the  extremities  of  the  diameter  of  a  circular  garden 
stands  two  trees,  one  20  feet  high  and  the  other  30  feet  high.  At 
what  point  on  the  circumference  must  a  ladder  be  placed  so  that 
without  moving  it  at  the  base  it  will  reach  to  the  top  of  each  tree, 
the  diameter  of  the  garden  being  40  feet. 

Construction. — Let  ABC  be  the  circular  garden  and  A  C  its 
diameter,  E  nd  let  AF  and  CD  be  the  two  trees  at  the  extremi- 
ties of  the  diameter.  Connect  the  tops  of  the  trees  by  the  line 
FD  and  from  the  middie  point  E  of  FD  let  fall  the  the  per- 
pendicular EH.  Draw  EG  perpendicular  to  FD.  Then  all 
points  of  EG  are  equally  distant  from  FD.  At  G  draw  BG 
perpendicular  to  AC.  Then  all  points  of  BG  are  equally  dis- 
tant from  F  and  D.  Hence,  B  is  the  required  point. 

1.  A  C=2/?=40ft.,  the  diameter  of  the  garden. 

2.  C0=«=30  ft.,  the  height  of  the  tree   CD,  and 

3.  AF=.--b=Z(j  ft,  the  height  of  the  tree  AF. 


II.  < 


—40  ft.— 30  ft.=lO  ft. 

5.  EH--=- 1  (  CD+  AF)  =4  (a+b) 

=£(40  ft+30  ft.)  =35  ft.      By 
similar  triangles, 

6.  FI\  ID  : :  EH-.  HG,  or 

Whence, 

7.  HG=C^^=m  ft 


FIG.  74. 


4— (*2— <*2)2-] 

-167?—  = 


ft. 


=34.91165  ft.,  nearly,  and 


320 


FINKEL'S  SOLUTION   BOOK. 


10.  ^c= 


82  f,=11.31942  ft. 


34.91165  ft.  the  distance  from  the  smaller  tree,  and 
11.31942  ft.  the  distance  from  the  larger  tree. 

I.  Seven  men  bought  a  grindstone  5  feet  in  diameter  ;  what 
part  of  the  diameter  must  each  grind  oft'  so  that  they  may  share 
equally? 

Construction. — Let  AB  be  the  diameter  of  the  grind  stone, 
O  its  center,  and  A  O  its* radius.  From  A  draw  any  indefinite 
line  AW  and  on  it  lay  off  any  convenient  unit  of  length  seven 
times,  beginning  at  A.  Let/* 
be  the  last  point  of  division. 
Draw  OP ',  and  from  the  other 
points  of  division  draw  lines 
parallel  to  OP,  intersect- 
ing the  radius  AO,  in  the 
points  /,  e,  d,  c,  6,  and  a.  Then 
the  radius  is  divided  into 
seven  equal  parts.  On  radius 
AO,  as  a  diameter  describe  a 
semi^circumference  A  OK,and 
at  a,  b,  c,  d,  e,  and/",  erect  per- 
pendiculars intersecting  the 
semi-circuro^ence  in  M,  L, 
K,  7,  H,  ana  G.  Then  with 
O  as  a  center  and  radii  equal  the  chords  MO,  LO,  KO,  7(9,. 
HO,  and  GO,  describe  the  circles  as  shown  in  the  figure.  Then 
each  man's  share  will  be  the  area  lying  between  the  circumfer- 
ences of  these  circles.  For,  the  chord  GO2=Gf2-\-fO2  and,  by 
a  property  of  the  circle,  Gf2=AfxfO.  .'.  GO^^AfxfO+fO*, 
=(Af-\-fO)fO=:A  OxfO=\A  O2.  In  like  manner  HO*=A  O 
X  £  0=4^4  O2,  KO2=^AOxdO=^AO2,  &c. 


1.  AB=D=5  ft.,  the  diameter  of  the  grind  stone. 

2.  A  O=R=1%  ft.,  the  radius. 

3.  .'.  rt/?2=7rx(2!)2=6^7r=the  area  of  the  stone. 

4.  |of7T/?2:=i7r^2=|of  6i7r==ff7r==each  man's  share. 

5.  Q^Tf — |f  7Tr=i|^r=the  area  of  the  stone  after  the  first  has 

ground  ofT  his  share. 

6.  .-.  V(i|7r^-7r)=1\y'42=2.31455+ft.,  the  radius  MO. 

7.  2(AO— MO)=2(2%  ft— T\V42  ft.)  =2 (2|  ft.— 2.31455 

ft)=.3709  ft,  part  of  the  diameter  the  first  grinds  off. 

8.  Q^TT — 2  Ofgj7r==yff5^===the  area  after  the  second  grinds 

off  his  share. 


MENSURATION. 


321 


9.  .-.  V(W7r-^7r)=f\/7==2-112875    ft.,   the    radius   LO. 
Then 

10.  2(MO—  £0)=2(5v/TV- f\/f  )=2(2.31455  ft— 

2.112875  ft)=.40335  ft,  the  part  of  the  diameter  the. 
second  grinds  off. 

11.  6i?r — f  cf  6i7T=2T57r=the     area     after    the     third    has 

ground  off  his  share. 

12.  .-.  V(3T57r-7-7r)=|\/4=5V'y=1.889822-4-  ft.,  the  radius 
"KG.     Then, 

1.8899822  tt)=.44(j  106  ft,  the  part   of  the  diameter 
the  third  grinds  oflf 

14.  61  n — %  of  §\it=\\ 7T==t'ie    area    after    the     fourth  has 

ground  off  his  his  share. 

15.  .-.  V(ll7r-T-7r)=-IVf  =  L636634ft'theradius/0'  Then 

16.  2(^0— 70)=2(5Vv-M)= 2(1.889822  ft.— 

1.636634  ft)=.50e!,!76  ft,the  part  of  the  diameter  the 
fourth  grinds  off. 

'.  §\7t—  f  of  6^=11*^116    area    after  the  fifth  grinds 
off  his  share. 


18. 


. 
-*-n=*  =  1.336306   ft,   the  radius    HO. 


Then 


20. 


«A,/_^Ffl     zv  7)=2(1.636634ft- 

1.336306  ft)=.600f«5/i  ft.,  the  part  of  the   diameter  the 

fifth  grinds  off. 
6^_e.0f  ^7r=|ft^=the  area   after    the   sixth    grinds 

off  his  share. 

21.  .•.»/(Hir-r-Jr)==|VH'^49911  ft.,  the  radius  GO.    Then 

22.  2(^0-GO)=2(iv/f-fV!)=2(  1.336306  ft— 

.944911  ft.)=.78i790  ft,  the  part  of  the  diameter  the 
sixth  grinds  off . 

23.  2  X  .944911  ft.=l. 889822  ft.,   the   diameter   of  the    part 

belonging  to  the  seventh  man. 

A   M.,  having  a  woolen  ball  2  feet  in  diameter,  bored  a 
hole  1  foot  in  diamer  through  the  center.     What    is    the   volume 

™ Construction. —Lz\.ABCDEF  be  a  great  circle  of  the  ball 
-and  let  A  CDF  be  a  verticle  section  of  the 
auger  hole.  Draw  the  diameter  BOE 
.and  the  radius  AC.  Then  the  volume 
bored  out  consists  of  a  cylinder,  of  which 
A  CDF  is  a  vertical  section,  and  two 
spherical  segments,  of  which  ACB  and 

~  are  vertical  sections. 

'1.  BE=%    feet=2/?,  the    radius    of 

ther  ball,  and 
9.  AO—-\  foot=2r,  the  radius  of  the 

auger  hole.  FIG.  76 


FINKEL'S  SOLUTION  BOOK. 


II. 


.  AF=2ij(R2 

V=nrz 


3. 

4.  . 

5. 

6.  2V/= 


7. 


=V3,  the  length  of  the  cylinder. 

the  volume  of  the  cylinder,  and 


.         r(l— |\/3)3=^7r(16-— 9\/3),   the  vol- 
ume ot  the  two  spherical  segments. 
.   F"+2F/=i^\/3-hV7r(16— 9\/3)=:^7r(8— 3>/3), 
=1.46809  cu.  ft.=2536.85952cu.  in. 

III.     .-.  The  volume  bored  out  is  2536.85952  cu.  in. 

I.     What  is  the  diameter  of  the  largest  circular    ring  that  cam 
be  put  in  a  cubical  box  whose  edge  is  1  foot? 

Construction. — Let  AB  CD — E  be  the  cubical  box.  Let  7,  K+ 
L,  M,  N,  and  P,  be  the  middle  points  of  the  edge  CF,  GF, 
HA,  AB,  and  BC  respec- 
tively. Connect  these  points 
by  the  lines  KI,  KL,  LM, 
MN,  NP,  and  PI.  Then 
IKLMNP  is  a  regular 
hexagon,  and  the  largest 
ring  that  can  be  put  in  the 
box  will  be  the  inscribed  cri» 
cle  of  the  hexagon. 


edge  of  the  cube. 


FIG.  77. 
Then 


TT  . 

AN\/2=$\S2e,  the 

side  of  the  hexagon, 

4.  MQ=ltML=ko 

5.  OR= 

the  radius  of  the  circle. 

6.  .-. 

14.6969382  in.,  the  diameter. 

III.  .•.  The  diameter  of  the  largest  circular  ring  that  can  be 
put  in  a  cubical  box  whose  edge  is  1  foot,  is  14.6969382  in. 

I.  A  fly  takes  the  shortest  route  from  a  lower  to  the  opposite 
upper  corner  of  a  room  18  feet  long,  16  feet  wide,  and  8  feet 
high.  Find  the  distance  the  fly  travels  and  locate  the  point 
where  the  fly  leaves  the  floor. 


MENSURATION.  323 

Construction.  —  Let  FABE  —  D  be  the  room,  of  which  AB  is 
the  length,  AF  the  width,  and 
AD  the  height;  and  let  F  be  the 
position  of  the  fly,  and  C  the  op- 
posite upper  corner  to  which  it 
is  to  travel.  Conceive  the  side 
ABCD  to  revolve  about  AB 
until  it  comes  to  a  level  with  the 
floor  and  takes  the  position  of 
ABC'D'.  Then  the  shortest 
path  of  the  fly  is  the  diagonal 
FC'  of  the  rectangle  FD'C'E, 
and  P  will  be  the  point  where  FIG.  78. 

the  fly  leaves  the  floor. 

1.  AB=a=.\%  ft,  the  length  of  the  room, 

2.  AF=b=l§  ft.,  the  width,  and 
A.<]3.  AD=A=&  ft.,  the  height. 

4.  ^ZX=^+^ZX=J+>&=16ft.+8ft.=24ft.    Then 
.5.  FC'=\/(FD"*+D'C'*)=  <t/[(£-j-/Oa+«8]» 
IL<  =V[(16+8)2-fl82]=30  feet,  the  length  of  the 

path  of  the  fly. 

l.  FD':D'C'  -.'.AF-.AP,   from    the    similar    triangles 
C'D'F  and  PAF,  or 


2.   b+h    :*::*:  AP.  Whence,  AP= 

o-\-/i 

==12  feet,  the  distance    from  A  to  where   the   fly 
leaves  the  floor. 

J30  feet  is  the  distance  the  fly  travels,  and  [floor. 

'  '1l2  feet  is  the  distance  from  A  to  where  it  leaves  the 

Remark.  —  If  we  conceive  the  side  BCHE  to  revolve  about 
EH  until  it  is  level  with  the  floor,  the  path  of  the  fly  will  be 
FC"  and  the  length  of  this  is  ^[(a+/i)*+t>*].  But  \f[(a+k)* 
-|-£2]  ^>  \/[(£-|-/z)2-|-tf2],  because,  by  expanding  the  terms  under 
the  radicals,  it  will  be  seen  that  the  terms  are  the  same,  except 
'Zah  and  2£^,  and  since  a  is  greater  than  £,  FC'  is  less  than  FC"  '. 
When  a=b,  FC'=FC". 

I.  How  many  acres  are  there  in  a  square  tract  of  land  con- 
taining as  many  acres  as  there  are  boards  in  the  fence  inclosing 
it,  if  the  boards  are  11  feet  long  and  the  fence  is  4  boards  high? 

1.  *  --  —  =mumber  of  acres  in  the  tract,  the  side  being  ex- 

pressed in  rods. 

2.  4  Xl6£X-^'<^—  number  of  feet  in   the   perimeter    of  the 

field. 


324  P^INKEL'S  SOLUTION  BOOK. 


3.  V  4X  riXl6^X5^g"|=number  of  boards  in  the  fence 

inclosing  the  tract. 

4.  ...  (£.a==4lxlXJlWg=24x^.     Whence, 


5. 

6.  .-.  «Vfe=3840  rods=12  miles. 
I?.  .'.  (3840)2-r-160=92160=number  of  acres  in  the  tract. 

Ill,     .'.  There  are  92160  A.  in  the  tract. 

(Milne's  Pract.  Arith.,  p.  362,  prob.  71.) 


SECOND   SOLUTION. 

1.  16=number  of  acres  comprised  between  two   panels   of 

fence  on  opposite  sides  of  the  field. 
2    lA.=43560sq.  ft. 
3.  16  A.=16X43560  sq.  ft.=696960  sq.  ft. 

llft.=the  width  of  this  strip  comprised  between  the  two 

panels. 

...  12  mi. =63360  ft=696960-KLl,  the  length  of  the  strip, 
which  is  the  width  of  the  field. 

6.  144  sq.  mi.=(  12 )2=the  area  of  the  field. 

7.  1  sq.  mi.=640  A. 

8.  144  sq.  mi.=144x640A=92160A. 

III.  .-.  There  are  92160  A.  in  the  tract. 

Explanation — Since  for  every  board  in  the  fence  there  is  an 
acre  of  land  in  the  tract  for  4  boards,  or  one  panel  of  fence  there 
would  be  4  A.  Now  a  panel  on  the  opposite  side  of  the  field 
would  also  indicate  4  A.  Hence,  between  two  panels  on  oppo- 
site sides  of  the  field  there  would  be  comprised  a  tract  11  ft.  wide 
and  containing  8  A.  But  this  would  make  boards  on  the  other 
two  sides  of  the  field  have  no  value.  Now  the  boards  on  the 
other  two  sides  having  as  much  value  as  the  boards  on  the  first 
two  sides,  it  follows  that  we  must  take  twice  the  area  of  the 
rectangle  included  between  two  opposite  panels  for  the  area  com- 
prised between  two  opposite  panels  in  the  entire  tract.  Hence, 
between  two  opposite  panels  in  the  tract  there  are  comprised  16  A. 
The  length  of  this  rectangle  is  16 X43560-f-1 1=63360  ft  =12  mi., 
which  is  the  length  of  the  side  of  the  tract. 

In  any  problem  of  this  kind,  we  may  find  the  length  of  a  side 
in  miles,  by  multiplying  the  number  of  boards  in  the  height  of 
the  fence  by  33  and  divide  the  product  by  the  length  of  a  board, 
expressed  in  feet. 


MENSURATION. 


325 


How  many  acres  in  a  circular  tract  of  land,  containing  as 
many  acres  as  there  are  boards  in  the  fence  inclosing  it,  the  fence 
being  5  boards  high,  the  boards  8  feet  long,  and  bending  to  the 
arc  of  a  circle? 

Construction. — Let  C  be  the  center  of  of  the  circular  tract,  AB 
=AC=1?,  the  radius,  and  the  arc  AjE>=8  feet.  Then  the  area  of 
.the  sector  is  5  A.=2 17800  sq.  ft. 

'1.  5A.=5x43560sq.  ft.=217800  sq.  ft,    the   area    of  the 
sector  ABC. 

2.  ^(A^XAC)=^(SX^C)=4AC=ai-ea     of     the     sector 

ABC. 

3.  /.  44C=217800sq.  ft.     Whence, 

4.  4C=217800-5-4=54450  ft.=3300  rods,  the  radius  of  the 

circle. 

5.  I-.   7t  X  (3300) 2-f-  160=68062.5*  =  number  of  acres    in 

tract. 

II.  .-.  There  are  68062.5  TT   A.,  in   the  tract. 

I.  What  is  the  length  of  a  thread  wrapped  spirally  around  a 
•cylinder  40  feet  high  and  2  feet  in  diameter,  the  thread  passing 
around  10  times? 

1.  27rft.=ABCA  (Pig.  19),  the  circumference  of  the  cyl- 
inder 
II J  2.  4  ft.=40  ft.-T-10=^/?i,  the  distance  between   the  spires. 

3.  \/[(27r)2+42]=2\/[>2+4]  .ft.=AJ£ir,   the    length    of 

one  spire. 

4.  .-.  10X2\/I>2+4]ft.  =20V|>8+4]  ft.=74.4838  ft,  the 

^entire  length  of  the  thread. 

III.  .-.  The  entire  length  of  the  thread=74.4838  ft 

Remark. — Each  spire  is  equivalent  to  the  hypotenuse  of  a  right 
angled  triangle  whose  base  is  ABC  A  and  altitude  A  P.  This 
may  be  clearly  shown  by  covering  a  cylinder  with  paper  and 
tracing  the  position  of  the  thread  upon  it  Then  cut  the  paper 
along  the  line  APK  o.r\d  spread  it  upon  a  plane  surface.  AEP 
will  then  be  seen  to  be  the  hypotenuse  of  a  right-angled  triangle 
whose  base  is  A  CBA  and  altitude  AP. 

I.  A  thread  passes  spirally  around  a  cylinder  10  feet  high 
and  1  foot  in  diameter.  How  far  will  a  mouse  travel  in  unwind- 
ing the  thread  if  the  distance  between  the  coils  is  1  foot  ? 

Construction. — Let  A  CB — /if  be  a  portion  of  the  cylinder  and 
ADEFGKv.  portioa  of  the  thread.  Let  A  be  the  position  of 
the  mouse  when  the  unwinding  begins,  P  its  position  at  any 
time  afterwards,  A  P N  a  portion  of  the  path  it  describes,  and  PD 
the  portion  of  the  thread  unwound.  DiawT  DC  parallel  *o  HB 
and  draw  QD  and  OC.  Then 


FINKEL'S  SOLUTION  BOOK. 


II. 


10. 


?=2/?=l  foot,    the    diameter    of    the    cylinder. 
0=10  ft.,  the  altitude.     Let 
#=the  angle  AOC, 
s=AW,  the  length  of  a  por- 
tion of  the  curve, 
x=0£,   and 
y=PL.     Then 


GM=R  cos  0, 

ML=fP=CP  cos  L   CPI 
=Rd   cos       7t—  l_  PCI) 


x=UM+ML=  7?  cos  0+R0 
sin  #,  and 

IM=  CM—  CI 
=R  sinV—CP  cos  0= 


dx=R6  cos  6  dB,  by  differ- 

entiating in  10, 

dy=R8s\n6d6,  by    differ- 

entiating in  11.  Now 


11.  r= 

12. 
13. 

14. 
15. 

(R0tiin0  dB)*}*=R  cede 

=\RQ*.     But 

16.  0=27T,    when    one    spire    is  unwound,  and 

17.  #=10x27T=20/T,  when  the  unwinding  is  complete. 

distance  the  mouse  travels  to  unwind  the  thread. 
III.  .-.  The  mouse  will  travel 989.96044  ft.  to  unwind  the  thread.. 

I.  What  is  the  length  of  a  thread  winding  spirally  round  a 
cone,  whose  radius  is  R  andaltitade  «,  the  thread  passing  round 
TZ  times  and  intersecting  the  slant  height  at  equal  distances  apart? 

Let  Ploe  any  point  of  the  thread,  (x,y,  z,)  the  co-ordinates  of 
the  point  ;and,  iet  the  angle  PFCf(=DOC)=0,  B0=a,  the  alti- 
tude, Z>0=£,  the  radius  of  the  base  of  the  cone,  and  r=the 
radius  of  the  cone  at  the  point  P.  Then  the  equations  of  the. 


thread  are  :  x-=~.r  cos/9  ......  (1),  y—r  sintf  .....  (2),  and  *= 

0  ____  (3).     From    the    similar    triangles    DBF   and   DOB* 

f  f) 

sYssv!?!    1  --     i  ...C4).    Now   the  distance  between 
V        'InnJ 

P  and  its  conoecutlvt   position  is 


— 
f? 


MENSURATION.  327 


zr....(5).       Substituting    the    value    of   r    in    (1) 
and    (2),    and    differentiating,    we 

have  dx— —  - — -     cos6-\- 
27rn\— 

(Inn — 6)  sin 'f\dO  and  dy=  — 


From  (3),  we  have  dz=-^^dO. 

Substituting   these    values    of   dx, 
dy,  and  dz  in  (5),  we  have 

Jo       2rc 


j?2 [sing — (2  n  n—B )cosV] 2  )        = 


where  7t=\/(a2-{-J?2) ,  the  slant  height. 

NOTE. — This  solution  was  prepared  for  the  School  Visitor,  by  the  author. 

I.  A  thread  makes  n  equidistant  spiral  turns  around  a  cone 
whose  slant  height  is  h,  and  radius  of  the  base  r.  The  cone 
stands  on  a  horizontal  plane  and  the  string  is  unwound  with  the 
lower  end  in  contact  with  the  plane,  the  part  unwound  being 
always  tense.  Find  the  length  of  the  trace  of  the  end  of  the 
string  on  the  plane. 

Let  MH\>e  the  part  unwound  at  any  time,  H  being  the  point 
in  contact  with  the  cone,  and  BM=u,  the  trace  on  the  plane  up 
to  this  time.  Put  arc  BE=x,  AH=y,  E  being  the  point  in  the 
circumference  of  the  base  in  the  line  AH.  Let  NI  be  the  posi- 


328  FINKEL'S  SOLUTION  BOOK. 

tion  of  the  string  at  the  next  instant,  D  and  /being  homologous 
points  with  E  and  H.         Draw  HK  parallel  to  ED.       Then  h  : 


DE  :  :  AK:  HK,  or== (1).     Now  since  the  arc  BE 


=x,  is  proportional  to  the  distance  the  point 
of  contact  of  the  thread  with  the  cone  has 

x         h — y 
ascended,  x'Ji — y\  '.%7trn'.h,  or- = — ~. 

*/  ' )  sV/lt  TT  A 


.•.  >-=-==. — tHHH (2).     This  is  negative 

since  y  decreases  as  x  increases.       It  is  evi- 
dent from  the  figure  that  TT^= — = - 

IK      dy  h 

By  similar  triangles,  IK'.HK \\  HE\ME, 

Jl/fT?          f-TR' 

that  is,  form  (1)  and  (2),  we  get- =~F^ 

y   dx^y 


Therefore,  ME= 

Put  ME=t. 

.  .  .  .  (5).     By  similar  figures  r\ME\  \ED\ 


^_ 


From  (3),  put  MP=», 

Equation  (5)  gives  the  entire   addition  to  the  line   ME   which 
-consists    of   NP+FD,    since    P*F=ME.       Consequently,  NP 
dt     dx         Znrn  %nrn     kitrn 

=lTy-dy=—h^(h-^+-ir=^ry  ....  (7).    Now  MN* 

=MP*+NP*  in  the  limit.     Therefore  ^lVy 


-^—(fi—y)2  )?  tne  intregal  of  which  is  «,  the   length  of 


z2.     Then  u=—  f  (7i—z) 
a^hJo    v 


the  trace.       Put  //—  •y=z1 


a  *  -f-  //  -  )  -i 

(11)-       Write   for  7z,  its  equal, 


MENSURATION.  329- 


4r       2^/^  2   ^\ 

««TT,  in  (11)  and  we  have  (12),  u=-  —  .-\-~inn  --  J 


This  result  is  independent  of  h,  the    cone's    slant   height,    but- 
involves  ;z  the  number  of  turns  of  the  thread. 

NOTE. — This  solution  is  by  Prof.  Henry  Gunder  and  is  taken  from  the 
School  Visitor,  Vol.9,  p.  199.  Prof.  Gunder  stands  in  the  very  front  rank 
V  Ohio  mathematicians.  He  has  contributed  some  very  fine  solutions  to 
difficult  problems  proposed  in  the  School  Visitor  and  the  Mathematical: 
Messenger.  He  is  of  a  very  retiring  disposition  and  does  not  make  any  pre- 
tentions  as  a  mathematician.  But  that  he  possesses  superior  ability  along 
that  line,  his  solutions  to  difficult  problems  will  attest.  Prof.  Gunder  was 
born  at  Arcanum,  O  ,  Sept.  15th,  1837.  He  passed  his  boyhood  on  a  farm 
and  it  was  while  following  a  plow  or  chopping  winter  wood,  that  difficult 
problems  were  solved  and  hitherto  unknown  fields  of  thought  explored.  He 
became  Principal  of  the  Greenville  High  School  in  1867.  After  seven  years* 
work  here,  he  became  Superintendent  of  the  Public,  Schools  of  North  Man- 
chester, Ind.  After  five  years'  work  at  this  place  he  became  Superintend- 
ent of  schools  of  New  Castle,  Ind  In  1890,  Prof.  Gunder  was  elected  pro- 
fessor of  Pedagogy  in  the  Findlay,  (Ohio)  College. 

I.  A  woman  printed  10  Ibs.  of  butter  in  the  shape  of  a  right 
cone  whose  base  is  8  inches  and  altitude  10  inches.  Having  com- 
pany for  dinner,  she  cut  offa  piece  parallel  to  the  altitude  and  con- 
taining \  of  the  diameter.  What  was  the  weight  of  the  part  cut  off?- 

Construction. — Let  ABC — G  be  the  cone,  A  C  the  diameter 
and  OG  the  altitude.  Let  E  be  the  point 
where  the  cutting  plane  intersected  the  the  di- 
ameter, F  the  corresponding  point  in  the  slant 
height,  and  DLFKB  the  section  formed  by  the 
intersection  of  the  cone  and  the  cutting  plane. 
Through  F  pass  a  plane  parallel  to  the  base 
AJBCand  anywhere  between  this  plane  and 

the  base,  pass  a  plane  NLMK.     Then, 

•!••• 

FIG-  82. 

1.  A  C=2^?=8in.,  the  diameter  of  the  base, 

2.  OG=a=lQ  in.,  the  altitude,  and 

3.  OE==OC—E  C=R—^A  C=R—IR=%R=\^  in.=c,the 

distance  of  the  cutting  plane  from  the    altitude.     Let 
4   GQ=x,   the  distance  of  the    plane  MLNK  from   the 
vertex  G.     Bv  similar  triangles, 

5.  OC\OG\\EC\EF,  or  R\a'.\R^c\EF.     Whence, 

6.  EF=a^     ~""^=6|  in.      By  similar  triangles, 

7.  GO:OC::GQ:QM,  or  a\R\\x\QM.     Whence, 

>=— .     Now, 
a 


330  FINKEL'S  SOLUTION  BOOK. 


II, 


9.  area  of  LKM=arca  of  LQ  KM—  area  of  LKQ.       But 
10.  area  of  LQKM=l(%LQ*  cosr1^^^^)1  X 


cos"'  and 

11. 


7?2V2  V^/rrX 

12.  ,'.  ^r^  of  the  segment  LKM==~--~--co*-*l  ^-}— 


ment  of  volume  of  the  part  cut  off. 


m 

14.  .-.  K=    I  (^r 

Jk 


=         19.6938154— 10.0562976 +  .6396202     = 


34.223792  cu.  in.,  the  volume  of  the  part  cut  off. 

15.  £tf7r^=iXlOx42X*==53£*r  cu.  in.,    the   volume  of 

the  whole  cone. 

16.  10  lbs.=the  weight  of  the  whole  cone.     Hence,  by  pro- 

portion, 

17.  53£7r  cu.  in. :  34.223792  cu.  in. : :  10  Ibs. :  (  ?=2.04258  Ibs. ) 

III.     /.  The  weight  of  the  part  cut  off  is  2.04258  Ibs. 

I.  After  making  a  circular  excavation  10  feet  deep  and  6  feet 
in  diameter,  it  was  found  necessary  to  move  the  center  3  feet  to 
one  side;  the  new  excavation  being  made  in  the  form  of  a  right 
cone  having  its  base  6  feet  in  diameter  and  its  apex  in  the  surface 
of  the  ground'  Reqired  the  total  amount  of  earth  removed. 


MENSURATION. 


331 


Construction, — Let  ABC — F  be  the  cylindrical  excavation 
first  made,  A  C  the  diame- 
ter, HO  the  altitude.  Let 
A  be  the  center  of  the  con- 
ical excavation,  GAH  its 
diameter,  and  AP,  an  ele- 
ment of  the  cylinder,  the 
altitude.  Pass  a  plane  at 
a  distance  x  from  O  and 
parallel  to  the  base  of  the 
excavation.  Let  figure  II. 
represent  the  section  thus 
formed,  the  letters  in  this 

section  corresponding  to  FIG.  88. 

the  homologous  points  in  the  base  represented  by  the  same  let- 
ters in  the  base  of  the  excavation.  An  element  of  the  earth  re- 
moved in  the  conical  excavation  is  (area  BAKGNB^dx.  The 
whole  volume  removed  in  the  conical  part  of  the  excavation  is 


r 

Jo 


(area  BAKGNB)dx.     For  let 


1.  HO  =  a=lQ  ft.,  the  altitude  of  the  excavation, 

2.  HA=r=3>  ft.,  the   radius  of   the   cylindrical   and    the 

conical  parts. 


This  is  found  from  the  proportion  of 

-^-a-\-AI).       Also 
£  ~~(2r—AI)AI=(rx-i-a— Af)(rx+a+Af).  Whence, 


3          A  /? A  7\71, 
.    ^~L  j-j n.  J.  v . 

a 
similar  triangles. 

4    /?/2 /  rx-^-a 

^    ^f2-—/2r 


r=r—J--=r(  l—~-^  }.        Now 


10.  area  of   BDAKGNB=1(  area    of  BDAN-\-area   of 

A^4G).       But 

11.  ^7r(r*x2-i-a2)=  the  area  of  the  quadrant  A^G,   and 

12.  area  of  BDAN=area  of  sector  BAN-\-area  of  trian- 

gle HBA—area  of  sector  BDAH.     Now 

13.  «r£0  of  sector  ^^A^^^^X^^sin-^^/-^-^^) 


—  «*),  and 


14.  area  of  triangle  ABH= 

XVC^8—  *f)=(ri* 

15.  area  of  sector  BDAH= 


332  FINKEL'S  SOLUTION  BOOK 

16.  .-.  Area  of  BDAKGNB=C1  \  \1-^L1t4-'  "*  sin"* 


o-£> 

S^2V2  ~2A 

r     *  ^""A 

— r2  cos-1^  1— ^1^  (  ^=i7r«r2+—  T^2  sin"3 

V      2«2y  i  '  «V« 


=the    volume    of    the    conical    part    of    the    exca- 
vation. 

18.  7tar2=ihe  volume  of  the  cylindrical  part. 

19.  ,. 


«r2=337.500554  cu.  ft.,  the  volume  of  the  entire  ex- 
cavation. 

III.  /.  The  volume  of  the  excavation  = 


or  337.50055-f  cu.  ft,  correct  to  the  last  decimal  place. 

NOTE.  —  This  problem  was  proposed  in  the  School  Visitor  by  Wayland* 
Bowling,  Rome  Center,  Mich.  A  solution  of  the  problem,  by  Henry  Gun- 
der,  was  published  in  Vol.  9,  No.  6,  p.  121.  The  solution  there  given  is  by 
polar  coordinates.  The  editor'gives  the  answers  obtained  by  the  contribu- 
tors; viz.,  Mr.  Dowling,  H.  A.  Wood,  R.  A.  Leisy,  and  William  Hoover. 
Their  answers  differ  from  Mr.  Gunder's  and  from  each  other.  Mr.  Gun- 
der's  answer  is  337.5-f-cu.  ft.,  the  same  as  above.  There  is  a  similar  problem^ 
in  TodhunteSs  Integral  Calculus,  p.  190,prob.  29. 

I.  A  tree  74  feet  high,  standing  perpendicularly,  on  a  hill- 
side, was  broken  by  the  wind  but  not  severed,  and  the  top  fell  di- 
rectly down  the  hill,  striking  the  ground  34  feet  from  the  root  of 
the  tree,  the  horizontal  distance  from  the  root  to  the  broken 
part  being  18  feet,  find  the  height  of  the  stub. 

Construction.  —  Let  AD  be  the  hill-side,  AB  the  stump, 


MENSURATION.  333 

the  broken  part,  and  A  C  the  horizontal  line  from  the  root  of  the 
tree  to  the  broken  part.  Produce  AB  to  E  and  draw  DE  paral- 
lel to  A  C. 

1.  Let  AB=x,  the  height  of  the  stump.     Then 

2.  BD=1±  ft. — x=s — #,  the  broken  part,  since  AB-\-BD 

=74  feet. 

3.  Let  AD=a=M  ft.,  the  distance  from  the    foot  of  the 

tree  to  where  the    top  struck  the  ground, 

4.  A  C=£=18  ft.,  the  horizontal   distance   from    the  foot 

of  the  tree  to  the  broken  part. 

5.  x=AJ3,  the  height  of  the  stump.     Then 

6.  BC=^(AB'2-\-A  C2)=V(*2+^2)  •  •  (1).    In  the  sim- 

lar  triangles      BA  C  and  BED, 

7.  ^/(#2_|_£2 )  .  x  .  .  s_x  .  BE.     Whence, 


II. 


11.  ^^ 


=«2  .  .  .  «(5).     Developing  (5),  we  have 

14.    4 


15.  1161^4—  91908*3-f  1959876*2—  25894080*+  377913600 

=0  .....  (7),  by  substituting    the  values   of  a,   b, 
and  s  in  (6). 

16.  .'.  x  =24  feet,  the  height  of  the  stump,  by   solving  (7) 

by  Homer's  method. 

III.     .-.  The  height  of  the  stump  is  24  feet. 

NOTE.  —  This  problem  was  taken  from  the  Mathematical  Magazine^  Vol. 
I.,  No.  7,  prob.  84.  In  Vol  I.,  p.  184,  of  the  Mathematical  Magazine  is  a  so- 
lution of  it,  given  by  C.  H.  Scharar  and  Prof.  J.  F.  W.  Sheffer  The  solu- 
tion there  given  is  different  from  the  one  above. 

I.  What  is  the  longest  strip  of  carpet  one  yard  wide  that  can 
be  laid  diagonally  in  a  room  30  feet  long  and  20  feet  wide  ? 

Construction.  —  Let  A  BCD  represent  the  room  and  EFGH 
the  strip  of  carpet  one  yard  wide  placed  diagonally  in  the  room. 


334  FINKEL'S  SOLUTION  BOOK. 


II. 


1.  Let  A£=a=3Q  ft.,  the  length  of  the  room, 

2.  £C=t>=20  ft.,  the  width,  and 

3.  HG=c=.%  ft.,  the  width  of  the  carpet.      Let 

4.  BF=HD=x.     Then 


6. 

7.  AE=GC=AB—EB=a— 

similar  triangles, 

8.  EF:BF::  GF:  GC,    or 

9.  c  :  x  :  :  GF:  a—>J(c2—x2.  ) 

Whence, 


X 

Again,  we  have 

11.  EF\BE\  :  GF'.FC,  or 

12.  c:<J(c*-x*)::GF:  b—  x.  FIGi  85. 


13.  .-.  LrJ?=-  .  .  .(4).  By  equating  G^in(3)  and(4), 

-x^ 


15.  I>x_xz==a^c2_x2)_c^x2  ...(6),  by  dividing  (5)  by 

cand  clearing  of  fractions. 

16.  c2  —  bx  —  2x2=a*J(c2—x2)  .  ..(7),  by  transposing  in(6). 

17.  4*4_4^3+(02+£2—  4c2)  x2+2tc2x=c*(a*—c*  )  .  .  .  - 

.  .  .  (8),  by  squaring  (7)  and    transposing    and    com- 
bining. 

18.  4*4—  80*  3+l  264*2+360*=  8019  ____  (9),  by  restoring 

numbers  in  (8). 

19.  .-.  #=.2.5571+ft,   by  solving  (9)    by  Horner's  method. 

20.  /V(^2—^2)=V(9—^2)—  1-5689  ft.      Then, 

21.  GC:=30—  V(9—  *2)—  28.4311  ft,  and 

22.  FC=  20—  *=17.4429  ft. 

23.  .-.  G:^^A/(^7^2+G:C2)=:V[(28.4311 

=33.3554  ft,  the  length  of.  the  carpet. 

III.     .'.     The  length  of  the  strip  of  carpet  is  33.3554  ft. 

I.  What  length  of  rope,  fastened  to  a  point  in  the  circumfer- 
ence of  a  circular  field  whose  area  is  one  acre,  will  allow  a  horse 
to  graze  upon  just  one  acre  outside  the  field  ? 

Construction.  —  Let  ABPC  be  the  circular  field  and  P  the 
point  in  the  circumference  to  which  the  horse  is  fastened.  Let 
BP  represent  the  length  of  the  required  rope.  Draw  the  radius 
BO  of  the  field  and  the  line  BC.  Then 

r  1.  1  A.=160  sq.  rd.=the  area  of  the  field  ABPC,  and 
2.  BO=OP=R=\/lW+n)=±\,   the  radius  of 


MENSURATION.  335 

the  circular  field.     Let 

3.  0=the  angle  ^/>O=the  angle  OB  P.     Hence, 

4.  TT—  2#=the  angle  BOP.     Now 

5.  BP=APcos£APB=ZRcosO,     the     length    of    the  - 

required  rope.     The 

6.  area  BP  CD  over  which  the  horse  grazes=area 

BE  CDB—areaBE  CPB. 
But 

7.  area  of  circle  BECD= 


47T./?2cos2#,  and  the 

area  BECP=^X(area  of 
sector  EPB-^-area  of  seg- 
ment BPH}.  Now 


arc  &JZ=±X*tfcosy  X  FIG,  86. 

2.ffcos9x0=2./?2flcos26>,  and 

10.  area  of  segment- BPH=area  of  sector  BOP — area 
triangle 
%[RXR(7t— 20)]—  1 

11. 


+  ^-2  6—  sin2  61  - 

20cos2fl—  siH22^]. 
12.  . 


sin2#].     But 

13.  7T/?2=1A.=160  sq.  rd.=the  area  of  BPCDB,    by  the 

conditions  of  the  problem. 

14.  .-.  47r7?2cos26>—  ^?2 

Whence, 


15.  4;r  —  7r2^cos2<9—sin2^=:7r   or 


16.  2?r--27rcos2#—  n— 

17.  . 

18.  2^  —  tan2#=27T,  by  dividing  by  cos2^.     Whence, 

19.  0=51°  16r  24X/,  by  solving   the    last   equation   by  the 

method  of  Double  Position. 

20.  .-. 

III.     .'.  The  length  of  the  rope  is  8.92926+rods. 

I.  If  a  2-inch  auger  hole  be  bored  diagonally  through  a  4-inch 
cube,  what  will  be  the  volume  bored  out,  the  'axis  of  the  auger 
hole  coinciding  with  the  diagonal  of  the  cube? 

Formula.  —  V=r2^(  ite  —  2r^2),  where  £  is  the  edge. 


336 


FINKEL'S  SOLUTION   BOOK. 


Construction.  —  Let  AFGD  be  the  cube  and  DF  the  diagonal,, 
which  is  also  the  axis  of  the  auger  hole.  The  volume  bored  out 
will  consist  of  two  equal  tetrahedrons  acd  —  D  and  efg  —  F  plus 
the  cylinder  acd—f,  minus  6  cylindrical  ungulas  each  equal  to 
ace  —  b.  Pass  a  plane  any  where  between  e  and  £,  perpendicular 
to  the  axis  of  the  cylinder,  and  let  x  be  the  distance  the  plane  is- 
from  D.  Now  let 


II. 


1.  AH—e=4:  inches,  the  edge  of  the  cube  ; 

2.  ZXF=\/3.y=4v'3,  the  diagonal  of  the  cube;  and 

3.  r=\  inch,  the    ra- 

dius of  the  auger, 
or  the  radius  of 
the  circle  acd. 

4.  ac—ad=  dc  =  r\/3 
=V3, 

5.  /te  =  iry6  =  iV6, 

by  the  similar  tri- 
angles dDc  and 
HDc. 


9. 
10. 

11. 


13. 
14. 


5=iV2,  the  al- 
titude of  the  tetra- 
hedron acd — D. 

.'.  2v  =  \(area  of  base  X  altitude)  = 
-^r\/2-)=-|y'6r3=-^Y'6,     the   volume    of  the  two  tetrahe- 
drons, 

v/=7fr2X(^^—2  times  the  altitude  of  acd—D)= 
7rr2(e\/3--i>V2)=7r(4A/3-— -h/2),  the  volume  of  the 
cylinder  acd — /". 

fo=^r\/2,  by  similar  triangles,  not  shown  in  the  figure. 

^r\/2-f-^r\/2=r\/2=distance  from    D    to    where    the 
auger  begins  to  cut  an  entire  circle. 

r — %x\f1=versine  of  an  arc  of  the  ungulas  at  a  distance 
x  from  D. 

12.  2rcos~~1f  •  y=an  arc  of  the  ungulas  at    a  distance 


x  from 
r2cos~1 


ment  at  a  distance  #  from  D. 


-i/*E*V 


MENSURATION. 


337 


15          V,the  volume  bored  c,ut,=2^+^/—  6^x/= 

—  7TV/2  )  =  r  V3  (  ^  — 


=16.866105  cu.  in. 
III.     .'.  The  volume  bored  out  is  16.866105  cu.  in. 
I.     A  horse  is  tethered  to  the  outside  of  a  circular  corral.     The 
length  of  the  tether  is  equal  to  the  circumference  of  the  corral. 
Required  the  radius  of  the  corral  supposing  the  horse  to  have  the 
libertj  of  grazing  an  acre  of  grass. 

Construction.—  Let  AEFBK  be  the  circular  corral,  AB  the 
diameter,  and  A  the 
point  where  the 
horse  is  tethered. 
Suppose  the  horse 
winds  the  tether 
around  the  entire 
corral;  he  will  then 
be  at  A.  If  he  un- 
winds the  tether, 
keeping  it  stretched, 
he  will  describe  an 
involute,  APGH\ 
to  the  corral.  From 
H'  to  H,  he  will  de- 
scribe a  semi-circle, 
radius  AH'=AH= 
to  the  circumference 
of  the  corral.  From 
H  through  G  to  A9 
he  will  again  de- 
scribe an  involute. 

Then  the  area  over  which  he  grazes  is  the  semi-circle  HLH'-\- 
the  two  equal  involute  areas  AFGHA  and  AKGH'A-\-\he  area 
BFGKB. 

Let  C  be  the  center  of  the  corral  and  also  the  origin  of  co-ordi- 
nates, A  G  the  *-axis  and  P  any  point  in  the  curve  APGH'. 

1.  Let  #=the  angle  A  CE  that  the  radius  CE  perpendic- 
ular to  PE,  the  radius  of  curvature  of  the  curve 
APGH'  ,  makes  with  the  #-axis, 

<90=the    angle   AFEBK  that  the   radius    CK  makes 
with  the  #-axis  when  the  radius  of  curvature  PE  has 
moved  to  the  position  KG\ 
R—A  C,  the  radius  of  the  corral; 


FIG.  88. 


2. 


3. 


4.  p=PE=arc  AFE=RV,   the   radius   of  curvature   of 


338  FINKEL'S  SOLUTION  BOOK. 

the  involute  ; 

5.  *=C3/and 

6.  y=PM,  the  co-ordinates  of  the  point  P\  and 

8.  y0=0,  the  co-ordinates  of  the  point  G.     Then  we  have 

9.  x=CM=lE—CD=PE(=arcAFJ5)  cosZ/^P, 

=RS  cos(  i  IEP—i  ECD}—R  cos(ar— 0)=7?0  cos 
[^TT — (n — 6)] — R  cos(7r—6)=ft6  cos  -(%n-—  8) 
—Rcos(7t—6)=R8cos8+R  sin0  . . . .  (1). 

10.  y=PM=PI+IM(=DE}=PE  sin^:  PEI+ECx 

sin  AECJD=RO  sin(# — ^7t)-\-R  sin(?r — ft)=R  s'mtf 
—RVsinV.  .  .  .  (2).      When  ^=^0=angle  AFEBK 

11.  *0=CG=^costf0+.##0sin6>0....  (3),  and 

12.  y0=0=R  sinV0—R60cosV0 (4).      Hence,  from   (4), 

13.  00=/?sin#0-:-/?cos#0=:tantf0 (5).    Then,  from  (3) r 

IL')-        =Rcos0 


(6).      Now 
15.  BFGKB=V(\KGKKC—  sector 


16. 

=±R*(%7i—ei)....  (8),  and 

17.  HH'L^\n(AHy=\7t(flnRY=:ln^R^  ....(9).     Ad- 

ding (7),  (8),  and  (9), 

18.  /?26io_/?2((9._7r)+^2(87ri_^3)+27r^2=: 

7?2(7r+L47T3—  ^^t)3)=area     over     which     the     horse 
grazes. 

19.  1  A.  =160  sq.  rd*=43560  sq.  ft.=the  area   over  which. 

the  horse  grazes. 

20.  .-.  /?2(7T+y7r3—  ^^)=43560  sq.  ft.      Whence, 


22.  B 0-=4.494039=2640  37X  18x/.35   by   solving  (5)  by   the 

method  of  Double  Position. 
V23    ;•.  7?=  19  24738  ft,  by  substituting  the  value  of  60  in(10). 

III.     .-.  The  radius  of  the  corral  is  19-24738  ft. 

A  20-foot  pole  stands  plumb  against  a  perpendicular  wall.  A 
cat  starts  to  climb  the  pole,  but  for  each  foot  it  ascends  the  pole 
slides  one  foot  from  the  wall;  so  that  when  the  top  of  the  pole  is 
reached,  the  pole  is  on  the  ground  at  right  angles  to  the  wall. 
Required  the  equation  to  the  curve  the  cat  described  and  the 
distance  through  which  it  traveled. 


MENSURATION. 


Construction.  =Let  A  C  be  the  wall,  P  the  position  of  the  cat 
at  any  time,  and  BC  the  position  of  the  ladder  at  the  same  time. 
Draw  AP  and  to  the  middle  point  D  of  AP  draw  BD.  Then 


1.  Let  #C=20  ft=0,  the  length  of  the  ladder, 

2.  AP=r,the    radius   vector   of  the   _____ 

curve  the  cat  describes,  and 

3.  0=the  angle  PAB. 

4.  TT — 20=the    angle  ABP,  because 

the  angle  PAB=\he  angle 
BPA. 


FIG.  89. 


II. 


III. 


6.  AP=%r=AD=ABcos^  BAD 

=  —  a  cos2#cos#. 

7.  .-.  r=  —  2«cos2#cos0,  or 

8.  r-}-2tf  cos2#cos#=0,   the  equation  of  the  curve    de- 

scribed by  the  cat. 

1.  Let  5=the  distance  through  which  the  cat  traveled. 

2.  s=J\/(dr*+r*de*)=ZaJ*  *  ^(1—  12  cos* 

44  cos4  6—  32  cos6  0)d6, 


f\p 

.  =  —  a  I 

Jo 

where  0 

/» 
.  =  —  £0  / 

.70 


cos0  —  cos20-|-4cos30)</0. 


where  0=;r  —  20, 


cos0  —  2 


=1.1193  a, 

5.  =22.386    ft,    the   distance    through    which  the   cat 
travels. 

r_|_2#cos2  #cos  0=0,  is  the  equation  or  the  curve,  and 
22.386  ft.=the  distance  through  which  cat  traveled. 


NOTE.  —  The  integration  in  this  problem  is  performed  by  Cotes'  Method 
of  Approximation. 

I.  Suppose  W.  A.  Snyder  builds  a  coke  oven  on  a  circular 
bottom  10  feet  in  diameter.  While  building  it,  he  keeps  one  end 
of  a  pole  10  feet  long,  always  against  the  place  he  is  working 
and  the  other  end  in  that  point  of  the  circumference  of  the  bot- 
tom opposite  him.  Required  the  capacity  of  the  oven. 

Construction.  —  Let  AB  be  the  diameter  of  the  base  and  CG 
the  altitude.  At  a  distance  x  from  the  base  pass  a  plane  inter- 
secting the  oven  in  /^and  E.  Draw  AE  and  A  C. 


340 


II. 


FINKEL'S  SOLUTION  BOOK. 

f  1.   AB={1R=\Q  feet,  the  diameter  of  the  base. 

2.  A  C=AJ5—<2fi=lQ  feet,  by  conditions  of  the  problem 

3.  CG=\/(AC2—  AG2)=(4:R2—  Ri)  =  R*]Z,   the    alti 
tude. 

4. 


=ZR*—2fiX  GH(=EI),   because 
the  ordinate  of  a  semi-circle    whose   diameter  is  2, 
From  this,  we  find 

/?.     Then 
—  A:2)—  ^?]2,  the 


area  of  the  circle  whose  center  is/. 

/rffys 
.-.    lr=  7 

Jo 

/ 

Jo 


Q        


FIG  90. 


9.  =^7T/?3(9V3— 47T)=:^53(9V3— 47T), 

10.  =i*rl25(9V3— 4?r)  =395.590202+  cu.  ft. 
III.     .-.   The  capacity  is  395.590202  cu.   ft. 

I.  At  each  corner  of  a  square  field  whose  sides  are  10  rods,  a 
horse  is  tied  with  a  rope  10  rods  long;  what  is  the  area  of  th» 
part  common  to  the  four  horses? 

Construction.— Let  ABCD  be  the  field  and  EFGH  the 
common  to  the  four  horses.  Join  EF, 
FG,  GH,  and  EH.  Draw  DKyzr 
peudicular  to  EF  and  draw  DE  and 
DF.  Since  AF=DE=£>F=  GB  = 
CE,  the  triangles  ADF  and  EDC 
are  equilateral  and,  consequently,  the 
angle  A  DE=  /  A  D  C—  Z  ED  C  =  90° 
— 60°=30°.  Also  the  angle  FDC= 
30C.  Hence,  EDF=3>&.  Now  let 


II. 


1.  AD=ED=a=\Q  rods. 

2.  Area  of  sector 


Then 


FIG.  91. 


arc 


3.   Area  of  uiangle  EDF=\EF^DK.     But, 
4. 


XXII.,  = 
6. 


2—^3),  and 


of  triangle 


,  by     formula     of    Prob. 


Hence, 

XW(2—  V3)=  i*: 


MENSURATION.  341 


7.  /.  Area  of  segment  EF=^7ta'i — ^a2=^a2(?r — 3).  The 

8.  area  of  square  EjFGH=EI?'i—a2 (2 — y'3).     Hence, 

9.  area  of  the  figure  EFGH=a'i(^—  y'3)+4x  ^a2(7t—  3) 

=«2(^7r-[-l — V3)=31.5147  sq.  rd.— the  area     common__ 
to  the  four  horses. 

III.     .-.  The  area  of  the   part  common   to    the  four  horses    is 
31.5147  sq.  rd. 

NOTE. — This  problem  is  similar  to  problem  348,  School  Visitor,  to  which 
a  fine  trigonometrical  solution  is  given  by  Prof.  E.  B.  Seitz. 

I.  What  is  the  length  of  the  longest  straight,  inflexible  stick 
of  wood  that  can  be  thrust  up  a  chimney,  the  arch  being  4  feet 
high  and  2  feet  from  the  arch  to  the  back  of  the  chimney — the 
back  of  the  chimney  being  perpendicular? 

Construction. — Let  PDE  C  be  a  verticle  section  of  the  chim- 
ney, PB  the  height  of  the  arch,  PE  the  distance  from  the  arch 
to  the  back  of  chimney,  and  APD  the  longest  stick  of  wood  that 
can  be  thrust  up  the  chimney. 

1.  Let    PjB=a—4:    feet,    the    height    of   the    arch, 

2.  PE=b=2  feet,  the  width  of  the 

chimney. 

3.  *=the  length  of  the  longest  stick 

of  wood,  and 

4.  fethe  angle  DA  C.     Then 

5.  AP=PB  cosec0=«  cosectf, 

7.  .'.  x=AP+PD=a  cosectf  +b 

sec  6 (1).        Differentia- 

_ting(l), 

II.,  cos4!...(2)!nor 

9.  a  cos*0=Z>sin*0  ....(3),  by  FIG.  92. 

clearing  of  fractions  and  transposing  in  (2). 

10.  .'.  ?^!==tan30=?.     Whence, 


11.  tan  #=;*]-.     From  (3),  we  may  also  have 

12.  cot#  =!H •     Now,  from  trigonometry, 

13.  v/(l_|-tan20  )=sec0,  and 

14.  v/(l-fcot2/9)=cosec^-     Hence,   by  substituting  in  (1), 

15.  x—ai/ 


342  FJNKEL'S  SOLUTION  BOOK. 

III.     .-  The  length  of  the  longest  stick  is  8.323876+ft 

I.  A  small  garden,  situated  in  a  level  plane  is  surrounded  by 
a  wall  having  twelve  equal  sides,  in  the  center  of  which  are 
twelve  gates.  Through  these  and  from  the  center  of  the  garden 
12  paths  lead  ofFthrough  the  plane  in  a  straight  direction.  From 
a  point  in  the  path  leading  north  and  at  a  distance  of  4  furlongs 
47_i^  yards  from  the  center  of  the  garden,  A.  and  B.  start  to  travel 
in  opposite  directions  and  at  the  same  rate.  A.  continues  in  the 
direction  he  first  takes;  B.,  after  arriving  at  the  first  road  (lying 
east  of  him)  by  a  straight  line  and  at  right  angles  with  it,  turns  so 
as  to  arrive  at  the  next  path  by  a  straight  line  and  at  right  angles 
with  it  and  so  on  in  like  manner  until  he  arrives  at  the  same 
road  from  which  he  started,  having  made  a  complete  revolution 
around  the  center  of  the  garden.  At  the  moment  that  B.  has 
performed  the  revolution,  how  far  will  A.  and  B.  be  apart? 

Let  O  be  the  center  of  the  garden,  A  the  point  in  the  path 
leading  north  from  which  A.  and  B.  start,  C,  D,  E,  F,  G,  H,  f, 
K,L,  M,  N,  P,  the  points  at  which 
B.  strikes  the  paths.  The  triangles 
OCA,  ODC,  OED,  OPE,  &c.,  are 
right  triangles,  OCA,  ODC,  OED, 
OEF,  &c.,  being  the  right  angles.  Let 
S  in  the  prolongation  of  A  C  denote  the 
position  of  A.,  when  B.,  arrives  at  P. 
It  is  required  to  find  the  distance  AS. 
Let  OA=a=4  furlongs,  47^¥\  yd->  Ps 
=A  C+  CD+DE+  .  .  .  +NP=x,  AS 
=y,  AP=z,  #=12,  the  number  of  paths 
and  LAOC=  LCOD=  LDOE=  ____ 
/AW)=360°-:-72=300.  Then  from  the  _ 
right  triangles  we  have  OC=OAX  FIG,  93. 


OD=OC  cosCO7}=a  cos2 
cos  DOE=a  cos3#,    OP=ONcos  NOP=a  cos^;  A  C= 
sinAOC=asin6,  C£>=OCsinCOZ)=a  sin0 


.-.  z=OA  —  OP--=a(\— 
tfsin(9  cos2  #-(-  .....  -\- 
cos3#-f-  .....  -(-cos^1 
acot%0(l—cosn0).       Hence,   since  £PAS=(W°+e  ),    we  have 

X 


yd.,  nearly. 


NOTE  —  This  problem  was  proposed  in  the  School  Visitor,  by  Dr.  N.  R. 
Oliver,  Brampton,  Ontario.  The  above  elegant  solution  was  given  by  Prof. 
E  B.  Seitz,  and  was  published  in  the  School  Visitor,  Vol  3,  p.  36. 


MENSURATION.  34S 

I.  A  fox  is  80  rods  north  of  a  hound  and  runs  directly  east 
350  rods  before  being  overtaken.  How  far  will  the  hound  run 
before  catching  the  fox  if  he  runs  towards  the  fox  all  the  time, 
and  upon  a  level  plain? 

Construction. — Let  C  and  A  be  the  position  of  the  hound  and 
fox  at  the  start,  P  and  m  corresponding  positions  of  the  hound 
and  fox  any  time  during  the  chase,  and  P'  and  n  their  positions 
the  next  instant,  B  the  point  where  the  hound  catches  the  fox 
and  CPP' B  the  curve  described  by  the  hound.  Join  m  and  P, 
and  n  and  P'jthey  are  tangents  to  the  curve  at  P  and  P' .  Draw 
Pd  and  P'e  perpendicular  to  AB ,  mo  perpendicular  to  /"«,. 
and  P'p  perpendicular  to  Pd. 


2.  AB=b= 350  rds., 
4!  JBd^ 


6.  arc  CP=s, 

7.  curve  CPB=s^  and 

8.  r=ratio  of  the  FIG  94. 

hound's  rate  to  the  fox's.     Then  we  have 

10!  ed=P'p=dy, 

11.  PP'=ds, 

12.  no—PP'=diu (1),  and 

13.  s—rx  ....  (2).     From  (2),  we  have,  by  differentiation,. 

14.  ds=rdx.  Whence, 

15.  -j-=~.    From  the  similar  right  triangles  PpP/  and  mon, 

we  have 

16.  PP/'.  ?nn\\pP/\  mo^  or  ds  ;  dx  '.'.  dy  :  mo.      Whence, 

_^ _dy    ^_^^dx_l 

11.1 

dy 

18.  -^-—ds—dw,  or 

r 

19.  ^y — r2^x==rc^w ^j       Integrating  (3), 

20.  y — r2x=rw-\-C. . .  .(4).       But,  since  when  #=0, 

and  w=a, 

21.  0=rfl+C.  Whence,  C=— 

22.  /.  y — rz x-=rw-\- C=rw — ra  ....  (5).     When  x=b,  j 

=^,and  w=0,  and  (5)  becomes 

23.  b — ^6= — ar,  or  r*b — ra=b.    Whence, 

94     ri ?, 1 

Zi^t.      /  —  ,'  1  5 


17.   mo=- — ; — — =— ,  since-T-=-.      Substituting:  in  (1). 
ds  r  ds     r 


344 


FINKEL'S  SOLUTION  BOOK. 


26- 

27. 


28.  r£=|(0-|-tf2+4£2).     But 

29.  rb=sl^  what  (1)  becomes  when  £=#. 

.30.  .*.  51=£(«+Vtf2-r-4£2y==;^2.2783  rods,  the  distance  the 
hound  runs  to  catch  the  fox. 


NOTE.  —  This  solution  is  substantially  the  same  as  the  one  given  by  the  Late 
Professor  E.  B.  Seitz,  and  published  in  the  School  Visitor,   Vol.2  V<  p.  201. 
The  path  of  the  hound  is  known  as  the  "Curve  of  Pursuit." 

I.  A  ship  starts  on  the  equator  and  travels  due  north-east  at 
^all  times  ;  how  far  has  it  traveled  when  its  longitude,  for  the 
first  time,  is  the  same  as  that  of  the  point  of  departure? 

Let  B  be  the  point  of  the  ship's  departure,  B  'PJVits  course,  P 
its  position  at  any  time  and  TV^its  position  at  the  next  instant. 
Then  PN  is  an  element  of  the  curve  of  the  ship,  which  is  known 
as  the  Loxodrome,  or  Rhumb  line.  Let  6=jBI?=\\\e  longitude 
of  the  point  P,  0=PJ7=the  corresponding  latitude,  (#,jy,  z) 
the  rectangular  co-ordinates  of  P  ,  and  ^.=^7r=the  constant 
angle  PNQ. 

Then  we  have  for  the 
equations  of  the    curve, 
x=PGcos6 
=rcos0cos#  ....  (1), 
y=P  G  sin  0=r  cos0  X 
sin#  ....  (2),  and 
z=r  sin0.  .  .  .(3),  where 
r  is    the    radius    of  the 
earth.    Now  an  element 
of  a     curve    of   double 
curvature,    referred    to 
rectangular  co-ordinates 

is  V(<fcM-< 

.-.  PN=ds 


...(4).     Differentiating 

<i),  (a),  £d  (*),.. 

cos0sin#df#),  f/G-  55. 

dy=  — r(sin#sin0</0 — cos0cos&/0),    and  dz=rcos  (f>d(f). 

stituting   these   values    in  (4),  ds=r<\/[cos'2(/)d0'2-\-(s'm 


Sub- 


.... (4).     Now 


and  N= 


Substituting  the  value  of  cos0^i9  i 


MENSURATION.  345- 


=ta.n(p,  or 


==tan9loga[tan(i?r+i0)]  or  ^cot(P=tan(i7r+|0)  .  .  .  (8). 
Whence,  0=2tan~1  (edcoi(P)—  \it  .     When  6^=2^  and  cp=l  ?r, 


the  distance  the  ship  travels. 

The  rectangular  equations  of  the  Loxodrome  are  V(^2 
j  £atan-i!_j_6ratair1!  (  =2r,  and  x2+y*+z2=r2  ,  where  «= 

The  last  equations  are  easily  obtained  from   the     figure.      The 


first  is  obtained  as  follows:  From  (1)  and  (2),  we  find 


v 
~J- 


X 


also,  #2  -f-J2=^2cos20  or  cos0=-V(^2+jK2).     From  (8),  we  get 

J^0  COS<^  Whence,  <0 «*<P+e<> «*<P * 

l+sin0     1+V(  I—cos2  0) 

os2  0)=cos0.    Transposing  eOcoW,  squaring,  and  reducing, 
we  have  cos0(^col<?>-f-^cot(?>)=2.     Substituting  the  value  cos0, 

(  y  y  ) 

and  #,  we  have  \/(x2-\-y^)  \  eai™~1K-\-e~ai&u~1v  >  =2r. 

NOTE.— This  solution  was  prepared  by  the  author  for  problem  1501> 
School  Visitor,  but  it  was  not  published  because  of  its  difficult  composi- 
tion. 


346  FINKEL'S  SOLUTION  BOOK. 


PARALLELOGRAMS. 

1.  Find  the  area  of  the  parallelogram  ABCD;  given  AC=7  ft.  2  in.,  and 
the  perpendicular  from  B  on  AC  3  feet.     [See  Fig.  4,  p.  198.] 

2.  Find  the  area  of  a  parallelogram  in  which  one  side  is  4  ft.  3  in.,  and 
the  perpendicular  distance  between  this  and  the  opposite  side  is  4  feet. 

3.  The  area  of  a  parallelogram  is  Yiy2  acres,  and  each  of  two  parallel 
sides  is  42  chains ;  find  the  perpendicular  distance  between  them. 

4.  Find  the  area  of  a  rhombus,  a  side  of  which  is  10  feet  and  a  diagonal 
of  12  feet.     [The  diagonals  of  a  rhombus  bisect  each  other  at  right  angles.] 

5.  Find  the  area  of  a  rhombus  whose  diagonals  measure  18  feet  and  24 
feet. 

6.  A  field  in  the  form  of  a  rhombus,  whose  diagonals  are  2870  links  and 
1850  links;  find  the  rent  of  the  field  at  $5  per  acre. 

7.  The  diagonals  of  a  parallelogram  are  34  feet  and  24  feet,  and  one  side 
is  25  feet ;  find  its  area. 

8.  Find  the  cost  of  carpeting  a  room,  30  feet  long  and  21  feet  wide,  with 
carpet  2  feet  wide  at  80  cents  per  yard. 

9.  How  many  square  yards  are  there  in  a  path,  4  feet  wide,  surrounding 
a  lawn  24  yards  long  and  22  yards  wide  ? 

10.  How  many  yards  of  paper,  20  inches  wide,  will  be  required  to  paper 
the  walls  of  a  room,  16  feet  by  14  feet  by  9  feet,  allowing  8  inches  for  a  base- 
board at  the  floor  and  12  inches  for  border  at  the  ceiling? 

11.  The  perimeter  of  a  rectangle  is  56  feet;  find  its  area,  if  its  length  is 
3  times  its  breadth. 

12.  What  is  the  area  in  acres  of  a  square  whose  perimeter  is  such  that 
it  takes  12  minutes  to  run  around  the  square,  at  the  rate  of  5)4  miles  per 
hour? 

13.  Cut  a  rectangular  board,  16  feet  long  and  9  feet  wide,  into  two  pieces 
in  such  a  way  that  they  will  form  a  square. 

14.  How  many  feet  of  framing,  4  inches  wide,  will  it  take  to  frame  a 
picture,  3  feet  by  2  feet  ?  Ans.  6  ft.  4  in. 

15.  A  sheet  of  galvanized  iron,  50  inches  wide,  is  placed  against  the  top 
of  a  wall,  6  feet  high,  while  the  lower  edge  is  5  feet  5  inches  from  the  foot 
of  the  wall ;  find  the  area  of  the  sheet  of  iron.  Ans.  4850  sq.  in. 

16.  Allowing  8  shingles  to  the  square  foot,  how  many  shingles  will  it 
take  to  roof  a  barn  which  is  40  feet  long  and  15  feet  from  the  comb  to  the 
eaves  ?  Ans.  9600  shingles. 

17.  The  area  of  a  square  is  169  sq.  ft. ;  find  its  perimeter,  in  chains. 

18.  What  is  the  side  of  a  square,  of  which  the  number  expressing  its 
area  in  square  feet  is  equal  to  the  number  expressing  its  perimeter  in 
yards?  Ans.  1^  feet. 

19.  What  is  the  area  of  a  path  a  yard  wide,  running  diagonally  across  a 
square  lawn  whose  side  is  30  feet?  Ans.  648[20l/~2— 1]  sq.  in. 

20.  What  is  the  area  of  a  square  whose  diagonal  is  12  feet  ? 

Ans.  72  sq.  ft. 

21.  What  is  the  area  of  a  square  whose  diagonal  is  5  feet  longer   than 
its  side?  Ans.  25 (3+2 /~3)  sq.  ft. 

TRIANGLES. 

1.  A  man  travels  20  miles  north,  then  15  miles  due  east,  finally  28  miles 
due  south ;  what  is  the  distance  from  his  starting  point  ?          Ans.  17  mi. 

2.  A  ladder,  50  feet  long,  is  placed  so  as  to  reach  a  window  48  feet  high, 


PROBLEMS. 


347 


and  on  turning  the  ladder  over  to  the  other  side  of  the  street,  it  reaches  a 
point  14  feet  high.     Find  the  breadth  of  the  street. 

3.  The  hypotenuse  of  a  right-angled  triangle  is  55  feet  and  the  base  is 
^  of  the  altitude.     Find  the  two  sides. 

4.  The  hypotenuse  of  a  right-angled  triangle  is  13  feet  and  the  sum  of 
the  sides  containing  the  right  angle  is  17  feet.     Find  these  sides. 

5.  In  a  right-angled  triangle  the  area  is  half  an  acre,  and  one  of  the 
sides  containing  the  right  angle  is  44  yards ;  find  the  other  side  in  yards. 

6.  Find  the  area  of  a  triangle  whose  sides  are  21  feet,  20  feet,  and  13 
feet,  respectively.     Also  21  feet,  17  feet,  and  10  feet. 

7.  In  a  right-angled  triangle  the  sides  containing  the  right  angle  are  30 
feet  and  40  feet.     Find  the  length  of  a  perpendicular  drawn  from  the  right 
angle  to  the  hypotenuse. 

8.  The  perimeter  of  a  triangle  is  48  feet.     If  one  side  is  10  feet  and  the 
area  is  84  square  feet,  find  the  two  remaining  sides. 

9.  The  area  of  an  equilateral  triangle  is  30  square  feet.     Find  the  length 
of  a  side.     What  is  the  side  of  a  square  of  equal  area  ? 

10.  The  sides  of  a  triangle  are  proportional  to  3,  4,  and  5.     If  the  perim- 
eter is  84  feet,  find  the  sides  and  the  area. 


TRAPEZOIDS. 

1.  The  parallel  sides  of  a  trapezoid  are  18  feet  and  24  feet,  and  the  alti- 
tude is  8  feet  ;  find  the  area. 

2.  The  parallel  sides  of  an  isosceles  trapezoid  are  16  feet  and  20  feet, 
and  the  non-parallel  sides  are  10  feet  each  ;  find  the  area  of  the  trapezoid. 

3.  The  line  joining  the  middle  points  of  the  non-parallel  sides  of  a 
trapezoid  is  12  feet,  and  the  altitude  is  8  feet  ;  find  the  area  of  the  trape- 
zoid. Ans,  96  sq.  ft. 

TRAPEZIUMS  AND   IRREGULAR   POLYGONS. 


1.  In  the  trapezium  ABCD,  AB=3&  in.,  BC=  17  in.,  CZ>=25  in., 

28  in.,  and  the  diagonal  £D=26  in.;  find  its  area.  Ans.  540  sq.  in. 

2.  In  the  quadrilateral  ABCD,  the  diagonal  AC=18  in.,  and  the  per- 
pendicular on  it  from  B  and  D  are  11  inches  and  9  inches  respectively; 
find  the  area  of  the  trapezoid.  Ans.  180  sq.  in. 

3.  In  the  trapezium  ABCD,  the  diagonals  AC  and  BD  are  perpendicu- 
lar to  each  other  and  measure  16  feet  and  2^  feet  respectively  ;  find  the 
area.  Ans.  2  sq.  yds. 

4.  Find  the  area  of  the  trapezium  ABCD,  in  which  the  angles  A£Cand 
CD  A  are  right  angles,  and  AB  is  15  feet,  BC  is  20  feet,  and  CD  is  7  feet. 

Ans.  234  sq.  ft. 

5.  Find  the  area  of  the  quadrilateral  ABCD,  having  given  that  the 
angle  ABC  is  60°,  ADC  is  a  right  angle,  ^=13  chains,  BC=IZ  chains, 
and  CD=12  chains.  Ans.  10.31  acres. 

6.  The  area  of  a  trapezium  is  4  acres  and  the  two  diagonals  measure 
16  chains  and  10  chains  respectively;  at  what  angle  are  the  two  diagonals 
inclined  to  each  other  ?  Ans.  30°. 

Hint.  —  Let  /  be  the  intersection  of  the  diagonals.    Then  1  from  B  on  AC=BI  sin. 
LBIC,\.  from  D  on  AC=DI  sin.^£>JA=(/_CIB.) 

.  •  .  area  of  ABCD=y2AC[BI  sin.  BIC+DI  sin.  CIB]=y2ACy.BDX$\n.  LBIC. 

7.  Find  the  area  of  the  polygon  ABCDEF,  if  AD=1G75  links,  1  FP 
from  .Ton  AD=8bQ  links,  1  BQ  from  B  on  ^Z?=-200  links,  1  CS  from  Con 
on  AD=50Q  links,  1  ER  from  E  on  AD=25Q  links,  AP=9QO  links,  AQ= 
1040  links,  AR=12W  links,  and  ^5=1380  links.  Ans.  9.03625  acres. 


348  'FINKEL'S  SOLUTION  BOOK. 

8.  Find  the  area  of  the  field  ABCDEF,  if  ^C=2900  links,  C£ 
links,  ^^=3600  liiiks,  1  BX  from  B  on  AC=4QQ  links,  i  £>Kfrom  D  on 
C£=4QQ  links,  and  1  FZ  from  F  on  ^£"=950  links.     Ans.  63  A.  3  r.  24  p. 

9.  Find  the  area  of  the  polygon  ABCDE,  if  AB=IZ  inches,  L^BC  a 
right  angle,  BC=§  in.,  CZ?=14  in  ,  AD=\5  in.,  /_ADE  a  right  angle,  and 
DJ5=8  in.  ^f«^.  1  sq.  ft.  30  sq.  in. 

REGULAR  POLYGONS. 

1.  Within  a  given  regular  hexagon,  drawn  on  a  side  of  10  inches,  a 
second  hexagon  is  inscribed  by  joining  the  middle  points  of  the  sides 
taken  in  order.   Find  the  area  of  the  inscribed  figure.  Am.  194.85  sq.  in. 

2.  Find  the  area  of  a  regular  pentagon  on  a  side  of  10  inches. 

Ans.  172.04  sq.  in. 

3.  Find  the  area  of  a  regular  decagon  on  a  side  of  4  inches. 

Ans.  123.1  sq.  in. 

4.  Find  the  area  of  a  regular  heptagon  inscribed  in  a  circle,  radius  6 

n  360° 

inches.    [Area  of  a  regular  #-side  in  terms  of  the  radius  is--  sin. 


5.  Find  the  area  of  a  regular  heptagon  circumscribing  a  circle  whose 
radius  is  12  inches.     [Area  of  a  regular  «-side  circumscribing  a  circle  in 

180° 
terms  of  the  radius  is  n  tan.-^—  (r)2]. 

6.  A  regular  octagon  is  formed  by  cutting  off  the  corners  of  a  square 
whose  side^£_12  inches.    Find  the  side  of  the  octagon.     [Side  of  octagoa 
=  i^a(2  —  v/  2),  where  a  is  the  side  of  the  square]. 

7.  The  area  of  a  dodecagon  is  300  square  inches  ;  find  the  radius  of  the 
circle  circumscribed  about  it. 

8.  Find  the  area  of  the  circular  ring  formed  by  the  inscribed  and  cir- 
cumscribed circles  of  a  regular  hexagon  whose  side  is  20  inches.     Show 
that  for  a  given  length  of  side,  the  area  of  the  ring  is  the  same  whatever 
the  number  of  sides  of  the  regular  polygons. 

9.  What  is  the  area  of  a  path  3  feet  wide,  around  a  hexagonal  enclosure 
whose  side  is  14  feet  ?  Ans.  283  .  17  sq.  ft 

10.  Find  the  area  of  the  square  formed  by  joining  the  middle  points  of 
the  alternate  sides  of  a  regular  octagon,  whose  side  is  8  inches. 

Ans.  186.51  sq.  in. 

11.  The  difference  between  the  area  of  a  regular  octagon  and  a  square 
inscribed  in  the  same  circle  is  82.8  square  inches.     Bind  the  radius  of  the 
circle.     [Take  j/~2=1.414].  Ans.  10  inches. 

12.  In  a  circle  of  a  radius  10  inches  a  regular  hexagon  is  described  ; 
in  this  hexagon  a  circle  is  inscribed  ;  in  this  circle  a  regular  hexagon  is 
inscribed  :  and  so  ad  infinitum.    Find  the  sum  of  the  areas  of  all  the  hex- 
agons thus  formed.  Ans.  1039.23  sq.  in. 

13.  In  the  last  example,  let  the  radius  be  r  and  the  number  of  sides  of 
the  polygon  n  ;  find  the  sum  of  the  areas  of  all  the  circles  formed. 

180° 

Ans.  K  r2  cosec.2  -- 
n 

14.  In  a  triangle  whose  base  is  15  inches  and  altitude  10  inches  a  square 
is  inscribed.    Find  its  area. 

CIRCLES. 

1.    The  driving-wheel  of  a  locomotive  engine  6  feet  3  inches  in  diam- 
eter, makes  110  revolutions  a  minute  ;  find  the  rate  at  which  it  is  traveling. 

Ans.  24.54  miles  per  hour. 


PROBLEMS. 


349 


2.  If  the  driving-wheel  of  a  bicycle  makes  560  revolutions  in  traveling 
a  mile,  what  is  its  radius?    [Take  *=%}].  Ans.  \VZ  feet. 

3.  Find  the  area  of  a  walk  7  feet  wide,  surrounding  a  circular  pond  252 
in  diameter.     [Take  *=3$].  Ans-  539°  sq-  ft- 

4.  A  wire  equal  to  the  radius  of  a  circle  is  bent  so  as  to  fit  the  circum- 
ference.    How  many  degrees  in  the  angle  formed  by  joining  its  ends  with 
the  center  of  the  circle  ?    [Take  * =3 . 14159265].  Ans.  57 . 2957795°. 

Definition. — The  angle  subtended  at  the  center  of  a  circle  by  an  arc 
equal  in  length  to  the  radius  is  called  a  radian. 

5.  A  wire  is  bent  into  the  form  of  a  circle  whose  radius  is  30  inches. 
If  the  same  wire  be  bent  into  the  form  of  a  square,  what  would  be  the 
length  of  its  side  ? 

6.  A  circle  and  a  square  have  the  same  perimeter.    What  is  the  differ- 
ence between  their  areas  ? 

7.  Two  tangents  drawn  from  an  external  point  to  a  circle  are  21  inches 
long  and  make  angles  with  each  other  of  90°.    Find  the  area  of  the  circle. 

8.  A  bicycle  driving-wheel  is  28  inches  in  diameter,  the  sprocket-wheel 
has  17  sprockets,  and  the  rear  sprocket-wheel  7  sprockets ;  what  is  the  gear 
of  the  wheel  ? 

Hint. — One  revolution  of  the  sprocket-wheel  makes  17-r-7=J^!!-  revolutions  of  the  rear 
sprocket-wheel,  or  JJZ.  revolutions  of  the  driving  wheel,  the  rear  sprocket-wheel  and  the 
driving-wheel  being  rigidly  connected.  .  •  .  7TX28X-L7-=7rX68  inches,  the  distance  trav- 
eled in  one  revolution  of  the  sprocket-wheel.  68  inches  is  the  gear  of  the  wheel.  Gear 
=  (  n-z-m  )£>,  where  «  is  the  number  of  sprockets  in  the  sprocket-wheel,  m  the  number  of 
sprockets  in  the  rear  sprocket-wheel,  and  D  the  diameter  of  the  driving-wheel  in  inches. 

9.  (a)  What  is  the  gear  of  a  bicycle  whose  driving-wheel  is  30  inches  in 
diameter,  whose  sprocket-wheel  has  19  sprockets,  and  whose  rear  sprocket- 
wheel  has  6  sprockets?     (b)  How  many  revolutions  of  the  sprocket-wheel 
will  be  required  to  travel  a  mile  ?  Ans.  (a)  95  inches. 

10.  What  is  the  distance  from  the  center  of  a  chord  70  inches  long  in  a 
circle  whose  radius  is  37  inches  ?  Ans.  12  inches. 

11.  In  a  circle  whose  radius  is  9  inches,  the  chord  of  half  an  arc  is  12 
inches ;  find  the  chord  of  the  whole  arc.  Ans.  17.89  inches. 

12.  The  length  of  an  arc  of  a  circle  is  143  inches  and  its  central  angle  is 
9°  6';  find  the  radius  of  the  circle.  Ans.  900  inches. 

13.  In  a  circle  of  a  radius  of  37  inches,  find  the  length  of  the  minor  arc 
whose  chord  is  24  inches.  Ans.  24.44  inches. 

14.  The  radius  of  a  circle  is  21  inches ;  find  the  length  of  an  arc  which 
subtends  an  angle  of  60°  at  the  center. 

15.  The  radius  of  a  circle  is  9  feet  4  inches ;  what  angle  is  subtended 
at  the  center  by  an  arc  of  28  inches  ? 

16.  The  chord  of  an  arc  is  48  inches  and  its  height  is  7  inches ;  find  the 
length  of  the  arc.     [Arc=^(8<5 — a),  where  b  is  the  chord  of  half  the  arc 
and  a  is  the  chord  of  the  whole  arc.]  Ans.  50%  inches. 

17.  In  a  circle  whose  diameter  is  72  inches,  find  the  length  of  the  arc 
whose  height  is  8  inches. 

18.  Find  the  area  of  a  sector  of  a  circle  whose  radius  is  21  inches  and 
the  angle  between  the  radii  40°. 

19.  Find  the  area  of  the  sector  of  a  circle  having  given  the  arc  32  inches 
and  the  radius  17  inches.  Ans.  272  sq.  in. 

20.  Angle  of  a  sector  is  36°  and  its  area  is  385  square  feet;  find  the 
length  of  its  arc.  Ans.  22  feet. 

21.  Find  the  area  of  a  segment  cut  off  by  a  chord  whose  length  is  14 
inches  from  a  circle  of  a  radius  of  25  inches.  Ans.  9.37  inches. 


350  FINKEL'S   SOLUTION   BOOK. 

32.*  A  regular  pentagon  is  inscribed  in  a  circle  of  a  radius  10  inches ; 
find  the  area  of  a  minor  segment  cut  off  from  the  circle  by  one  of  its  sides. 

Ans.  15.27  sq.  in. 

33.  Find  the  area  of  a  segment  whose  chord  is  30  inches  and  height  is 
8  inches.  Ans.  168.16  sq.  in. 

34.  Find  the  area  of  a  circle  inscribed  in  a  sector  whose  angle  is  120° 
and  whose  radius  is  10  inches. 

35.  A  line  AB  is  20  inches  long,  and  C  is  its  middle  point.    On  AB,  AC, 
and  CB  semicircles  are  described.    Find  the  area  of  the  circle  inscribed  in 
the  space  inclosed  by  the  three  semicircles.  Ans.  r=3^  inches. 

36.  Two  equal  circles,  each  of  a  radius  9  inches,  touch  each  other  exter- 
nally, and  a  common  tangent  (direct)  is  drawn  to  them ;  find  the  area  of  the 
space  inclosed  between  the  circles  and  the  tangent.  Ans.  7 . 53  sq.  inches. 

37.  Three  circles  of  radius  3  feet  are  placed  so  that  they  touch  each 
other ;  find  the  area  of  the  curvilinear  space  inclosed  by  them. 

Ans.  207  sq.  in. 

38.  From  the  angular  points  of  a  regular  hexagon,  whose  side  is  10 
inches,  six  equal  circles,  radii  5  inches,  are  drawn  ;  find  the  area  of  the 
figure  inclosed  between  the  circles.  Ans.  50(3V  3—^)  sq.  in. 

39.  Two  equal  circles  of  radius  5  inches  are  described  so  that  the  center 
of  each  is  on  the  circumference  of  the  other ;  find  the  area  of  the  curvi- 
linear figure  intercepted  between  the  two  circumferences. 

Ans.  30.71  sq.  in. 

40.  Two  equal  circles  of  radius  5  inches  intersect  so  that  their  com- 
mon chord  is  equal  to  their  radius ;  find  the  area  of  the  curvilinear  figure 
intercepted  between  the  two  circumferences.  Ans.  4.53  sq.  in. 

41.  Three  circles,  radii  10,  12,  and  16  inches  respectively,  touch  each 
other;  find  the  radius  of  a  circle  touching  the  three  circles.     [See  Prob. 
CLXVL] 

SIMILAR   AREAS. 

[See  principle  on  p.  309.] 

1.  The  sides  of  a  triangle  are  21,  20,  and  13  inches;  find  the  area  of  a 
similar  triangle  whose  sides  are  to  the  corresponding  sides  of  the  first 
as  25: 3. 

2.  In  a  survey  map  an  estate  of  144  acres  is  represented  by  a  quadrilat- 
eral, ABCD.     The  diagonal,  AC,  is  6  inches,  and  the  perpendiculars  from 
B  and  D  on  AC  are  1.8  inches  and  .9  inches  respectively.     On  what  scale 
was  the  map  drawn  ?  Ans.  6  inches  to  the  mile. 

3.  A  man  6  feet  in  height,  standing  15  feet  from  a  lamp-post,  observes 
that  his  shadow  cast  by  the  light  at  the  top  of  the  post  is  8  feet  in  length ; 
how  long  would  his  shadow  be  if  he  were  to  approach  8  feet  nearer  to  the 
post  ?  Ans.  2  ft.  4  in. 

4.  A  man,  wishing  to  ascertain  the  width  of  an  impassable  canal,  takes 
two  rods,  3  feet  and  5  feet  in  length.     The  shorter  he  fixes  vertically  on 
one  bank  and  then  retires  at  right  angles  to  the  canal,  until  on  resting  the 
other  rod  vertically  on  the  ground  he  sees  the  ends  of  the  two  rods  in  a 
line  with  the  remote  bank ;  if  the  distance  between  the  rods  is  60  feet, 
what  is  the  width  of  the  canal  ?  Ans.  90  feet. 

5.  A  man  wishing  to  find  the  height  of  a  tower,  fixes  a  rod  11  feet  in 
length  vertically  on  the  ground  at  a  distance  of  80  feet  from  the  tower. 
On  retiring  10  feet  further  from  the  tower  he  sees  the  top  of  the  rod  in 
line  with  the  top  of  the  tower.     If  the  observer's  eye  is  5^  feet  above  the 
ground,  find  the  height  of  the  tower.  Ans.  55  feet. 

6.  A  triangle  ABC  is  divided  into  two  equal  parts  by  a  straight  line 
XYt  drawn  parallel  to  the  base  BC.    If  A£=WO  inches,  find  AX. 


PROBLEMS.  351 

7.  In  a  given  triangle  a  triangle  is  inscribed  by  joining  the  middle 
points  of  the  sides.     In  this  inscribed  triangle  another  similar  triangle  is 
inscribed,  and  so  on.     What  fraction  of  the  given  triangle  is  the  area  of 
the  sixth  triangle  so  drawn  ? 

8.  (a)  In  a  given  square  whose  side  is  16  inches  a  square  is  inscribed  by 
joining  the  middle  points  of  the  sides  of  the  given  square  ;  in  this  inscribed 
square  a  square  is  inscribed  in  like  manner,  and  so  on  ;  find  the  area  of  the 
fifth  square,     (b)  If  the  process  be  continued  ad  infinitum  what  is  the  sum 
of  the  areas  of  all  the  squares  ? 

9.  In  a  circle  of  a  radius  of  32  inches  an  equilateral  triangle  is  in- 
scribed, and  in  this  triangle  a  circle.     In  this  circle  an  equilateral  triangle 
is  again  inscribed,  and  in  the  triangle  a  circle,  and  so  on.     If  the  process  is 
continued,  find  the  area  of  the  fourth  circle  and  find  which  of  the  circles 
has  an  area  of  3£  sq.  in.?  Ans.  50f  sq.  in.;  the  sixth. 

10.  A  field  of  9  acres  is  represented  in  a  plan  by  a  triangle  whose  sides 
are  25,  17,  and  12  inches.     On  what  scale  is  the  plan  drawn  and  what  length 
will  be  represented  by  80  inches?  Ans.  ,-^;  1  mile. 

12.  The  following  is  used  by  lumbermen  in  finding  the  diameter  of 
trees  at  any  height  above  the  ground  :  If  the  tree  casts  a  definite  shadow 
on  a  horizontal  plane,  stand  on  the  edge  of  the  shadow  and  observe  where 
the  line  of  light  from  the  sun  to  your  eye  strikes  the  tree.  Then  measure 
the  shadow  of  the  tree  at  the  point  where  the  shadow  of  your  head  strikes 
the  ground.  The  width  of  the  shadow  is  the  diameter  of  the  tree  at  the 
point  where  the  line  of  light  from  the  eye  to  the  sun  strikes  it.  What 
principle  is  involved  ? 


1.  Find  the  surface  of  a  rectangular  solid  whose  length  is  12  feet, 
breadth  5  feet  4  inches,  height  5  feet  3  inches. 

2.  Find  the  cost  of  papering  the  four  walls  of  a  room  whose  length  is 
20  feet  6  inches,  breadth  15  feet  6  inches,  and  height  11  feet  3  inches,  at  3d. 
a  square  yard. 

3.  A  rectangular  tank  is  16  feet  long,  8  feet  wide,  and  7  feet  deep  ;  how 
many  tons  of  water  will  it  hold,  a  cubic  foot  of  water  weighing  1,000  oz.? 

4.  The  surface  of  a  rectangle  is  1,000  sq.  in.;  if  its  length  and  breadth 
are  respectively  1  ft.  3  in.  and  1  ft.  2  in.,  find  its  height. 

5.  The  dimensions  of  a  rectangular  solid  are  proportional  to  3,  4,  and  5. 
If  the  whole  surface  contains  2,350  sq.  in.,  find  the  length,  breadth,  and 
height. 

/#«/.—  2,350-5-[2(3X4)  +2(3X5)  +2(4X5)]=25,  the  greatest  common  divisor  of  the  three 
dimensions. 

6.  The  whole  surface  of  a  rectangular  solid  contains  1,224  square  feet, 
and  the  four  vertical  faces  together  contain  744  square  feet.     If  the  height 
is  12  feet,  find  the  length  and  breadth. 

7.  Find  the  surface  and  volume  of  a  cube  whose  diagonal  is  2  feet  6 
inches.  Ans.  12^  sq.  ft.  ;  3  cu.  ft.  12  cu.  in.,  nearly. 

8.  Find  the  edge  of  a  cubical  block  of  lead  weighing  one  ton,  having 
iven  that  a  cubic  foot  of  lead  weighs  709^  Ibs.         Ans.  17  .60+  inches. 

9.  The  edges  of  a  rectangular  block  of  granite  are  proportional  to  2,  3, 
id  5,  and  its  volume  is  101  cu.  ft.  432  cu.  in.  ;  find  its  dimensions. 

Ans.  3  ft.  ;  4  ft.  6  in.  ;  7  ft.  6  in. 

10.  The  diagonal  of  a  rectangular  solid  is  29  inches,  and  its  volume  is 
4,032  cu.  in.  ;  if  the  thickness  is  one  foot,  find  the  length  and  breadth. 

Ans.  21  in.  and  16  in. 


given 

9. 
and  5 


352  FINKEL'S  SOLUTION  BOOK. 


PRISMS. 

1.  A  right  prism  stands  upon  a  triangular  base,  whose  sides  are  13,  14, 
and  15  inches.    If  the  height  is  10  inches,  find  its  volume  and  whole  sur- 
face. Ans.  840  cu.  in.  ;  4  sq.  ft.  12  sq.  in. 

2.  The  weight  of  a  brass  prism  standing  on  a  triangular  base  is  875  Ibs. 
If  the  sides  of  the  base  are  25  in.,  24  in.,  and  7  in.,  find  the  height  of  the 
prism,  supposing  that  1  cu.  ft.  of  brass  weighs  8,000  oz.  Ans.  3  ft. 

3.  Water  flows  at  the  rate  of  30  yards  per  minute  through  a  wooden 
pipe  whose  cross-section  is  a  square  on  a  side  of  4  inches.     How  long  will 
it  take  to  fill  a  cubical  cistern  whose  internal  edge  is  6  feet? 

Ans.  21f  min. 

4.  Find  the  volume  of  a  truncated  prism  (  that  is  the  part  of  a  prism 
included  between  the  base  and  a  section  made  by  a  plane  inclined  to  the 
base  and  cutting  all  the  lateral  edges),  whose  base  is  a  right  triangle,  base 
3  feet,  and  altitude  4  feet,  and  the  three  lateral  edges  3  feet,  4  feet,  and  5 
feet  respectively.    [Formula.  —  J/r=^A(e1L-{-es-\-e3)J  where  A  is  the  area  of 
the  base  and  elt  es,  and  es  the  lateral  edges.] 

CYLINDERS. 

1.  How  many  cubic  yards  of  earth  must  be  removed  in  constructing  a 
tunnel  100  yards  long,  whose  section  is  a  semi-circle  with  a  radius  of 
10  feet? 

2.  Find  the  convex  surface  of  a  cylinder  whose  height  is  three  times  its 
diameter,  and  whose  volume  is  539  cubic  ihches. 

3.  The  cylinder  of  a  common  pump  is  6  inches  in  diameter  ;  what  must 
be  the  beat  of  the  piston  if  8  beats  are  needed  to  raise  10  gallons? 

Ans.  12     in. 


4.  A  copper  wire  ^  inches  in  diameter  is  evenly  wound  about  a  cylin- 
der whose  length  is  6  inches  and  diameter  9.9  inches,  so  as  to  cover  the 
convex  surface.    Find  the  length  and  weight  of  the  wire,  if  1  cu.  in.  of 
copper  weighs  5.1  oz.  Ans.  1,885  in.,  nearly;  75.5  oz. 

5.  A  cubic  inch  of  gold  is  drawn  into  a  wire  1,000  yards  long.    Find  the 
diameter  of  the  wire.  Ans.  .006  in. 

6.  The  whole  surface  of  a  cylindrical  tube  is  264  square  inches  ;  if  its 
length  is  5  inches,  and  its  external  radius  is  4  inches,  find  its  thickness. 
[Use7r=3f.]  Ans.  I  in. 

7.  If  the  diameter  of  a  well  is  7  feet,  and  the  water  is  10  feet  deep,  how 
many  gallons  of  water  are  there,  reckoning  7^  gallons  to  the  cubic  foot? 

PYRAMIDS  AND  CONKS. 

1.  Find  the  entire  surface  of  a  right  pyramid,  of  which  the  height  is 
2  feet  and  the  base  a  square  on  a  side  of  1  ft.  8  in.  Ans.  10  sq.  ft. 

2.  Find  the  convex  surface  of  a  right  pyramid  1  foot  high,  standing  on 
a  rectangular  base  whose  length  is  5  feet  10  inches  and  breadth  10  inches. 

Ans.  8  sq.  ft.  128  sq.  in. 

3.  Find  the  con-vex  surface  of  a  right  pyramid  having  the  same  base 
and  height  as  a  cube  whose  edge  is  10  inches.  Ans.  223.6  sq.  in. 

4.  Find  the  weight  of  a  granite  pyramid  9  feet  high,  standing  on  a 
square  base  whose  side  is  3  feet  4  inches,  1  cubic  foot  of  granite  weighing 
165  Ibs.  Ans.  2  tons,  9  cwt.  12  Ibs. 

5.  Find  the  height  of  a  pyramid  of  which  the  volume  is  623  .  52  cu.  in., 
and  the  base  a  regular  hexagon  on  a  side  of  1  foot.  Ans.  5  inches. 


PROBLEMS.  353 

6.  The  volume  of  a  regular  octahedron  is  471.41  cubic  feet;  find  the 
length  of  each  edge.  Ans.  10  feet. 

7.  Find  the  surface  of  a  regular  tetrahedron,  if  the  perpendicular  from 
one  vertex  to  the  opposite  face  is  5  inches. 

8.  A  conical  vessel  is  5  inches  in  diameter  and  6  inches  deep.     To  what 
depth  will  a  ball  4  inches  in  diameter  sink  in  the  vessel  ? 

9.  The  ends  of  the  frustum  of  a  pyramid  are  squares  whose  sides  are 
20  inches  and  4  inches,  respectively.     If  its  altitude  is  15  inches,  what  is 
its  convex  surface  ?  Ans.  110  sq.  in. 

10.  What  is  the  volume  of  a  frustum  of  a  pyramid  whose  upper  base  is 
4  inches  square,  lower  base  28  inches,  and  the  length  of  the  slant  edges 
15  inches  ? 

11.  The  volume  of  a  frustum  of  a  cone  is  407  cubic  inches  and  its  thick- 
ness is  W}4  inches?    If  the  diameter  of  one  end  is  8  inches,  find  the  diam- 
eter of  the  other  end.     O=^.]  Ans.  6  inches. 

SPHERES. 

1.  Find  the  ratio  of  the  surface  of  a  sphere  to  the  surface :     (i)  of  its 
circumscribed  cylinder,  (ii)  of  its  circumscribed  cube. 

2.  A  cube  and  a  sphere  have  equal  surfaces ;  what  is  the  ratio  of  their 
volumes  ?  Ans.  72 : 100,  nearly. 

3.  From  a  cubical  block  of  rubber  the  largest  possible  rubber  ball  is 
<mt.     What  decimal  of  the  original  solid  is  cut  away? 

4.  Suppose  the  earth  to  be  a  perfect  sphere,  8,000  miles  in  diameter ;  to 
what  height  would  a  person  have  to  ascend  in  a  balloon  in  order  to  see 

2^* 
one-fourth  of  its  surface?    [Formula. — h—     2  »  where  r  is  the  radius  of 

the  earth,  and  —  is  the  part  of  the  earth's  surface  visible  to  the  observer. 

If  the  part  of  the  earth  visible  to  the  observer  is  .£-,  or  1/-2L,  «=-—• 

q  P  P 

5.  A  paring  an  inch  wide  is  cut  from  a  smooth,  round  orange  an  inch 
and  a  half  in  diameter.    What  is  its  volume,  if  it  is  cut  from  the  orange 
on  a  great  circle  of  the  orange  ?  Ans.  \"^. 

6.  What  would  be  the  volume  of  a  paring  cut  from  the  earth  on  the 
equator?  Ans.  $*aat  where  a  is  the  width  of  the  paring. 

Remark—  This  is  a  remarkable  fact,  since  the  volume  of  the  paring  is  independent 
of  the  radius  of  the  sphere. 

7.  If,  when  a  sphere  of  cork  floats  in  the  water,  the  height  of  the  sub- 
merged segment  is  #  of  the  radius,  show  that  the  weights  of  equal  vol- 
umes of  cork  and  water  are  as  34 :44. 

Note—  The  weight  of  a  floating  body  is  equal  to  the  weight  of  the  liquid  it  displaces. 

8.  A  vertical  cylindrical  vessel  whose  internal  diameter  is  4  feet,  is  com- 
pletely filled  with  water.     If  a  metal  sphere  25  inches  in  diameter  is  laid 
upon  the  rim  of  the  vessel,  find  what  weight  of  water  will  overflow. 

Ans.  699  Ibs.,  nearly. 

9.  A  conical  wine-glass  5>^V  3  inches  in  diameter  and  4  inches  deep  is 
filled  with  water.     If  a  metal  sphere  5'/£  inches  in  diameter  is  placed  in  the 
vessel,  what  fraction  of  the  whole  contents  will  overflow  ?  Ans.  \. 

10.  Four  equal  spheres  are  tangent  to  each  other.     What  is  the  radius 
of  a  sphere  tangent  to  each  ? 

11.  To  what  depth  will  a  sphere  of  ice,  three  feet  in  diameter,  sink  in 
water,  the  specific  gravity  of  ice  being  -f  ? 


354  FINKEL'S  SOLUTION  BOOK. 

12.  Find  the  volume  removed  by  boring  a  2-inch  auger-hole  through  a 
6-inch  globe. 

13.*  What  is  the  volume  removed  by  chiseling  a  hole  an  inch  square 
through  an  8-inch  globe  ? 

PRISMATOIDS  AND  WEDGES. 

1.  Find  the  weight  of  a  steel  wedge  whose  base  measures  8  inches  by  5 
inches,  and  the  height  of  the  wedge  being  6  inches ;  if  1  cu.  in.  of  steel 
weighs  4 . 53  oz.  ? 

2.  Find  the  volume  of  a  prismatoid  of  altitude  3.5  cm.,  the  bases  being 
rectangles  whose  corresponding  dimensions  are  3  cm.  by  2  cm.  and  3.5  cm. 
by  5  cm. 

3.  The  base  of  a  wedge  is  4  by  6,  the  altitude  is  5,  and  the  edge,  ^,  is  3. 
Find  the  volume. 

RINGS. 

1.  Find  the  surface  and  volume  of  a  ring,  the  radius  of  the  inner  cir- 
cumference being  10>£  inches  and  the  diameter  of  the  cross-section  3>£ 
inches.  Ans.  847  sq.  in. ;  741£  cu.  in. 

2.  Find  the  surface  and  volume  of  a  ring,  the  diameters  of  the  inner 
and  outer  circumferences  being  9.8  inches  and  12.6  inches  respectively. 

Ans.  154.88  sq.  in. ;  54.21  cu.  in. 

SIMILAR  SOLIDS. 

1.  The  edges  of  two  cubes  are  as  4:3;  find  the  ratio  of  their  surfaces 
and  their  volumes. 

2.  The  surfaces  of  two  spheres  are  in  the  ratio  of  25:4;  find  the  ratio 
of  their  volumes. 

3.  At  what  distance  from  the  base  must  a  cone,  whose  height  is  1  foot, 
be  cut  by  a  plane  parallel  to  the  base,  in  order  to  be  divided  into  two  parts 
of  equal  volume  ?  Ans.  2 . 47  in. 

4.  A  right  circular  cone  is  intersected  by  two  planes  parallel  to  the  base 
and  trisecting  the  height.     Compare  the  volumes  of  the  three  parts  into 
which  the  cone  is  divided.  Ans.  1:7:19. 


EXAMINATION  T^STS. 


ARITHMETIC. 

1.  How  do  you  divide  one  fraction  by  another?    Why  is  the  fraction 
thus  divided  ? 

2.  Divide   four  million   and   four   millionths  by  one  ten  -  thousandth. 
Write  the  answer  in  figures  and  words. 

3.  If  a  liter  of  air  weighs  1 . 273  gr.,  what  is  the  weight  in  kilos.,  if  the 
air  is  in  a  room  which  contains  78  cu.  m.  ? 

4.  The  base  of  a  cylinder  is  12  inches  in  diameter  and  its  altitude  is  25 
inches.     Required  the  solid  contents. 

5.  The  edge  of  a  cube  is  6  inches ;  what  is  the  length  of  the  diagonal 
of  the  cube  ? 

6.  A  broker  bought  stock  at  4%  discount,  and  sold  it  at  5%  premium^ 
and  gained  $450.     How  many  shares  did  he  purchase? 


PROBLEMS. 


355 


7.  A  ships  500  tons  of  cheese,  to  be  sold  at  9^  cents  a  Ib.    He  pays  his 
agent  3%  for  selling ;  the  proceeds  are  to  be  invested  in  sugar,  after  a  com- 
mission of  2%  is  deducted  for  buying.     Required  the  entire  commission. 

8.  Upon   what  value  are  dividends    declared?     Brokerage  estimated? 
Usual  rate  of  brokerage  ? 

9.  What  is  the  face  of  a  note  dated  July  5, 1881,  and  payable  in  4  months 
to  produce  $811,  when  discounted  at  9%? 

10.  Upon  what  principle  is  the  United  States  rule  for  partial  payments 
based?    The  Mercantile  rule?    How  does  compound  interest  differ  from 
annual  interest?  Ohio  State  List,  1884. 


For  the  benefit  of  students  preparing  for  county  or  state  examinations, 
we  write  out  the  answers  to  the  above  questions  as  a  specimen  of  how  the 
examination  paper  ought  to  be  prepared : 


SUBJECT  :    Arithmetic. 

Name  of  Applicant. 


(a)  Invert  the  terms  of  the  divisor  and  then  multiply  the 
numerators  of  the  fractions  together  for  the  numerator  of  the 
quotient  and  the  denominators  together  for  the  denominator  of 
the  quotient. 

(6)  The  fraction  is  thus  divided  because  inverting  the  terms 
of  the  divisor  gives  the  number  of  times  the  divisor  is  con- 
tained in  1,  as  is  shown  by  analysis.  The  number  of  times  then 
it  is  contained  in  any  other  number  is  obtained  by  multiplying 
this  number  by  the  number  of  times  the  divisor  is  contained 
in  1. 


II. 


1. 


Four  million  and  four  millionths=4000000. 000004= 
40000000000004 

1000000 

1 
One  ten-thousandth  =  .0001  = 


3. 


10000 

40000000000004    1    40000000000004   10000 

X  i 


1000000    '  10000 
40000000000004 


1000000 
4 


I 


100 


-=400000000000^=400000000000 . 04= 


III.     .*.  The  quotient  is  four  hundred  billion  and  four  huudredths. 


II. 


III. 


1.  1  cu.  m.=1000  1. 

2.  78  cu.  m. =78X1000  1=78000  1. 

3.  1.273  g.=the  weight  of  1  1.  of  air,  and 

4.  99294  g. =78000X1  -273  g.= weight  of  78000  1. 

5.  1000  g.=l  kilo. 

6.  99294  g. =99294  g. -=-1000=99. 294  kilos. 

/.  The  weight  of  78  cu.  m.  of  air  weighs  99.294  kilos. 


356 


FINKEL'S  SOLUTION  BOOK. 

Arithmetic —  Continued. 


12  in.=the  diameter  of  the  cylinder,  and 
25  in.=the  altitude.     Then 


II.  \  3. 
4. 


III. 


X7rl28=367r  sq-  in.,  the  area  of  the  base  of  the  cylinder. 
25X36^=900*  Cu.  in.=900X3.  141592X1  cu.  in.= 
2827.4328  cu.  in.,  the  volume  of  the  cylinder. 
/.  The  volume  of  the  cylinder  is  2827  .4328  cu.  in. 


1.  6  in.=the  length  of  the  edge  of  the  cube. 

2.  36  sq.  in.-|-36  sq.  in. =72  sq.  in. = the  area  of  the  square 

described  on  the  diagonal  of  one  of  the  equal  faces, 
which  is  the  sum  of  the  areas  of  the  squares  de- 
scribed on  two  equal  edges. 

**•  '  3.     72   sq.  in.-f-36   sq.  in. =108  sq.  in.=area  of  square  de- 
scribed on  the  diagonal  of  the  cube,  which  equals  the 
sum  of  the  areas  described  on  the  three  edges. 
4.     6V3ln.=vT08Xl  in. =10. 392+  in.,  the  length  of  the  diag- 
onal of  cube. 
III.     /.  6V3ln.=10.392+  in.=length  of  diagonal  of  cube. 


II. 


III. 


1.  100%= par  value  of  stock. 

2.  4%= discount. 

3.  96%  =  100%—  4%= market  value,  or  cost  of  stock. 

4.  5%  =  premium. 

5.  105%  =  100% -f5%= selling  price  of  stock. 

6.  9%  =  105%— 96%=gain. 

7.  $450=gain. 

8.  /.9%  =$450. 

9.  1%=$50,  and 

10.  100%=$5000=par  value  of  stock. 

11.  $100=par  value  of  one  share,  usually.     Then 
1 12.    $5000=par  value  of  $5000-r-$100,  or  5  shares. 

.'.  He  purchased  5  shares. 


II. 


10 


II. 


9j4  cents=selling  price  of  one  Ib. 
$47.50=500X$0.09^=selling  price  of  one  Ib. 
fl.     100%  =$17. 50. 
1% =$0.475. 

2%=2X$0.475=$0.95=commission  for  selling 
the  cheese. 
$47.50— $0  95=$46.55=proceeds,  or  the  amount  to  be 

invested  in  sugar. 
100%=cost  of  sugar. 

3%=commission  on  sugar. 
103%=total  cost  of  sugar. 
$46.55=total  cost  of  sugar, 
flv     .'.  103%  =$46. 55. 
j  2.  l%=Tfo  of  $46.55=$0.45. 

1  3.        100%=100X$0.45=$45=cost  of  sugar. 
14.  2%=2X$0.45=$0.90=commission  on  sugar. 

,    .'.  $0 . 95+$0 . 90=$1 . 85=total  commission. 
$1.85=entire  commission. 


PROBLEMS. 
Arithmetic —  Concluded. 


357 


Dividends  are  declared : 

(a)  Upon  the  par  value. 

(b)  Brokerage  is  reckoned  upon  the  selling  price  or  purchas- 
ing price  of  bonds  in  Commission  and  Brokerage,  but  in  Stock 
Investments  it  is  reckoned  on  the  par  value. 

(c)  The  usual  rate  of  brokerage  is  £%  on  the  par  value  of  the 
stock,  either  for  a  purchase  or  a  sale. 


100%=face  of  note. 
3^% ^discount  for  126  da. 
)6  ££% = proceeds. 
$811=proceeds. 


1.  100%=face  of  note.  1881—  7— 

2.  3A<^=discount  for  126  da.  4 
3. 

II.  -I  4. 
5. 
6. 
7. 
III.   " /.  The  face  of  the  note  must  be  $837.372. 


when  dated. 


1881—11—  5-8  when  due. 


1% =$8. 37372,  and 

=$837.372,  the  face  of  the  note. 


10 


(a)  Upon  the  principle  that  payments  be  applied  first  to  the 
discharge  of  interest  due,  the  balance,  if  any,  toward  paying  the 
principal  and  interest.    Interest  or  payment  must  in  no  case 
draw  interest. 

(b)  Upon  the  principle  that  partial  payments  shall  draw  inter- 
est from  time  of  payment  until  date  of  settlement. 

(c)  Compound  Interest  increases  in  a  geometrical  ratio,  and 
Annual  Interest  in  an  arithmetical  ratio. 


1.  A  and  B  together  have  $9,500.     Two-thirds  of  A's  money  equals  f  of 
B's.     How  much  money  has  each  ? 

2.  A  owes  a  sum  equal  to  f  of  his  yearly  income.     By  saving  TV  of  his 
income  annually  for  5  years,  he  can  pay  his  debt  and  have  $1,200  left.  What 
is  his  yearly  income? 

3.  Smith  and  Jones  can  do  a  piece  of  work  in  12  days.     If  Smith  can  do 
only  f  as  much  as  Jones,  how  long  will  it  take  each  of  them  to  do  the  work  ? 

4.  I  am  offered  6%  stock  at  84,  and  5%  stock  at  72.    Which  investment 
is  preferable,  and  how  much  ? 

5.  If  in  selling  cloth  f  of  the  gain  is  equal  to  ^  of  the  selling  price,  for 
how  much  will  3^  yards  sell  that  cost  $5  per  yard  ? 

6.  The  frustum  of  a  cone  is  10  feet  in  diameter  at  the  bottom  and  8  feet 
at  the  top,  with  a  slant  height  of  12  feet.     What  is  the  height  of  the  cone 
from  which  the  frustum  is  cut  ? 

7.  A,  B  and  C  ate  eight  pies.    If  they  ate  equal  shares  and  A  and  B  furnish 
the  pies,  and  C  pays  16  cents  for  his  share,  how  should  A  and  B  divide  the 
money? 

8.  Which  is  the  heavier,  and  how  much,  an  ounce  of  lead  or  an  ounce 
of  gold?  Pickaway  County  List,  f8oo. 


1.  Define  bonds,  coupons,  exchange,  tariff. 

2.  A  field  of  12  acres  and  30  perches  yields  255  bu.  2  qts  of  wheat;  how 
much  will  a  field  of  15  acres  and  10  perches  yield  at  the  same  rate? 

3.  Find  value  of  11%  of  $180  +  22%  of  $160  -f  92%  of  $63. 


358  FINKEL'S  SOLUTION  BOOK. 

4.  A  piano  was  sold  for  $297,  at  a  gain  of  35% ;  what  would  have  been 
the  %  of  gain  if  it  had  been  sold  for  $300  ? 

5.  A  dealer  imported  120  dozen  champagne,  invoiced  at  $23  a  dozen,, 
breakage  12£% ;  what  was  the  duty  at  22%  ? 

6.  I  rent  a  house  for  $300  per  year,  the  rent  to  be  paid  monthly  in  ad- 
vance ;  what  amount  of  cash  at  the  beginning  of  the  year  will  pay  one 
year's  rent  ?  . 

7.  The  rafters  of  a  house  are  20  feet  long,  the  width  of  the  gable  is  30= 
feet,  the  rafters  project  two  feet ;  what  is  the  height  of  the  gable  ? 

8.  What  the  convex  surface  of  the  frustum  of  a  cone  whose  slant  height 
is  6  feet,  the  diameter  of  its  lower  base  5  feet,  and  of  its  upper  4  feet? 

9.  To  be  analyzed  :    If  for  every  cow  a  farmer  keeps,  he  allows  \  acre  for 
pasture,  and  f  of  an  acre  for  corn,  how  many  cows  can  he  keep  on  39  acres  ? 

10.  How  much  can  I  afford  to  give  for  6's  of  '81  so  that  I  may  realize  8% 
per  annum,  gold'being  at  a  premium  of  15?  Hancock  County  List. 

1.  What  is  the  surface  of  a  parallelepiped,  8  feet  long,  4  feet  wide,  and 
2  feet  high? 

2.  A  starts  on  a  journey  at  the  rate  of  3  miles  per  hour ;  6  hours  after- 
wards B  starts  after  him  at  the  rate  of  4  miles  per  hour.     How  far  will  B 
travel  before  he  overtakes  A? 

3.  The  time  since  noon  is  ^  of  the  time  to  4  o'clock  P.  M. ;  what  is  the 
time? 

4.  A  man  having  oranges  at  4  cents  each,  and  apples  at  2  for  1  cent, 
gained  20%  by  selling  5  dozen  for  $2.04 ;  how  many  of  each  did  he  sell  ? 

5.  The  first  term  of  a  geometric  series  is  3,  the  third  term  507  ;  find  the 
ratio. 

6.  A  merchant  sold  a  quantity  of  goods  at  a  gain  of  20%.     If,  however, 
he  had  purchased  the  goods  for  $60  less,  his  gain  would  have  been  25%. 
What  did  the  goods  cost  ? 

7.  There  is  a  park  400  feet  square ;  a  walk  3  feet  wide  is  made  in  it,  along 
the  edges,  how  many  square  yards  would  such  a  walk  contain  ? 

8.  A  man  sold  wheat,  commission  3%  and  invested  the  proceeds  in  corn, 
commission  2%  his  whole  commission,  $250 ;  for  how  much  did  the  wheat 
sell  and  what  was  the  value  of  the  corn  ?  Licking  County  List. 

1.  A  man  had  43f  yards  of  carpeting,  costing  $26^ ;  he  sold  f  of  the 
pieces  gaining  %\  on  each  yard  sold.     How  much  did  he  receive  for  it? 

2.  From  the  product  of  f£   and  -f^    subtract  the  difference  of  their 
squares. 

3.  How  many  acres  in  a  field  whose  length  is  40  rods  and  diagonal  50 
rods? 

4.  How  many  trees  will  be  required  to  plant  the  above  by  placing  them 
1  rod  apart  ?    By  2  rods  apart  ? 

5.  Bought, a  lot  of  glass;  lost  15%  by  breakage.     At  what  %  above  cost 
must  I  sell  the  remainder  to  clear  20%  on  the  whole  ? 

6.  After  spending  25%  of  my  money,  and  25%  of  the  remainder,  I  had 
left  $675.     How  much  had  I  at  first  ? 

7.  How  many  fifths  in  ^?  Ans.  If. 

8.  A  box  is  3£  inches  long,  2£  inches  wide,  and  2  inches  deep  will  con- 
tain how  many  J-inch  cubes  ? 

9.  Change  f  of  quart  to  the  decimal  of  a  bushel. 

10.  A  can  hoe  16  rows  of  corn  in  a  day,  B  18,  C  20,  and  D  24.     What  is 
the  smallest  number  of  rows  that  will  keep  each  employed  an  exact  num- 
ber of  days  ?  Seneca  County  List. 


MENSURATION.  35£ 

1.  (a)     Define:    number,  integer,  fraction,  a  common  multiple,  and 
the  greatest  common  divisor  of  two  or  more  numbers. 

(b)  Prove  (do  not- merely  illustrate}  that  to  divide  by  a  fraction 
one  may  multiply  by  the  divisor  inverted. 

(c)  Change  74632  from  a  scale  of  8  to  a  scale  of  9. 

2.  (a)     The  freezing  and  boiling  temperatures  of  water  are  32°  and 
212°,  respectively,  when  measured  by  a  Fahrenheit  thermometer;    meas- 
ured by  a  centigrade  thermometer  they  are  0°  and  100°,  respectively;    if 
a  Fahrenheit  thermometer  records  a  temperature  of  74°  what  would  the 
centigrade  record  be  at  the  same  time? 

(b)     By  what  per  cent  must  8°  Fahrenheit  be  increased  so  as  to- 
equal  8°  centigrade? 

3.  Silver  weighs  10.45  times  as  heavy  as  water,  while  gold  weighs  19.30 
times  as  heavy  as  water;    find,  correct  to  3  decimal  places,  the  number 
of  inches  in  the  edge  of  a  cube  of  gold  which  is  equal  in  weight  to  a  cube 
of  silver  whose  edge  is 4.3  cm.     Also  express  this  weight  in  (Troy)  grains. 

4.  A  6%  bond,  which  matures  in  3  years,  with  interest  payable  annu- 
ally, is  selling  at  104 ;  a  §?%  bond,  which  matures  in  H  years,  with  interest 
payable  semi-annually,  is  selling  at  102.     Which  is  the  better  investment? 
And  how  much  better  is  it? 

5.  A  water-tank  has  connected  with  it  4  pipes;    the  first  can  fill  it  in 
30  min.,  the  second  in  40  min.,  the  third  can  empty  it  in  50  min.,  and 
the  fourth  can  empty  it  in  one  hour.     If  these  pipes  are  so  arranged  that 
the  third  is  automatically  opened  when  the  tank  is  precisely  £  filled,  and 
the  fourth  when  the  tank  is  f  filled,  how  long  will  it  take  to  just  fill  the 
tank  if  the  second  pipe  is  set  running  10  minutes  later  than  the  first? 

T. 
Cornell  University  —  Scholarship  Examination,  1899. 

1.  A  and  B  run  a  race,  their  rates  of  running  being  as  17  to  18.    A 
runs  2£  miles  in  16  minutes,  48  seconds  and  B  the  whole  distance  in  34 
minutes.     What  is  the  distance  run? 

2.  The  surface  of  the  six  equal  faces  of  a  cube  is  1350  sq.  inches. 
What  is  the  length  of  the  diagonal  of  the  cube? 

3.  A  man  bought  5%  stock  at  109 J,  and  4J%  pike  stock  at  107£,  broker- 
age in  each  case  £% ;    the  former  cost  him  $200  less  than  the  latter,  but 
yielded  the  same  income.     Find  the  cost  of  the  pike  stock. 

4.  A,  B,  and  C  start  together  and  walk  around  a  circle  in  the  same 
direction.     It  takes  A   ^  hours,   B  f  hours,   C   f-f-  hours  to  walk  once 
around  the  circle.  How  many  times  will  each  go  around  the  circle  before 
they  will  all  be  together  at  the  starting  point? 

5.  I  hold  two  notes,  each  due  in  two  years,  the  aggregate  face  value 
of  which  is  $1020.     By  discounting  both  at  5%,  one  by  bank,  the  other 
by  true  discount,  the  proceeds  will  be  $923.     Find  face  of  bank  note. 

6.  The  hour  and  minute  hands  of  a  watch  are  together  at  12  o'clock; 
when  are  they  together  again? 

7.  How  many  cannon  balls  12  inches  in  diameter  can  be  put  into  a 
cubical  vessel  4  feet  on  a  side ;    and  how  many  gallons  of  wine  will  it 
contain  after  it  is  filled  with  the  balls,  allowing  the  balls  to  be  hollow, 
the  hollow  being  6  inches  in  diameter,  and  the  opening  leading  to  it  con- 
taining one  cubic  inch? 

8.  An  agent  sold  a  house  at  2%  commission.     He  invested  the  pro- 
ceeds in  city  lots  at  3%  commission.     His  commissions  amounted  to  $350. 
For  what  was  the  house  sold? 

Ohio  State  List,  December,  1898. 


:360  FINKEVS  SOLUTION  BOOK. 

1.  A,  B,  and  C  can  do  a  piece  of  work  in  84  days;    A,  B,  and  D  in 
72  days;    A,  C,  and  D  in  63  days;    B,  C,  and  D  in  56  days.    In  what 
time  can  each  do  it  alone? 

2.  A  banker  bought  U.   S.  4's  at  128|%  and  U.   S.  4£'s  at  106^, 
brokerage  |%.     The  latter  cost  him  $1053.75  more  than  the  former,  but 
yielded  him  $195  more  income.    How  much  was  invested  in  each  kind 
of  bonds? 

3.  f  of  the  cost  of  A's  house  increased  by  £  of  the  cost  of  his  farm 
for  2  years  at  5%,  amounts  to  $4950.     What  was  the  cost  of  each,  if  f 
of  the  cost  of  the  house  was  only  f  as  much  as£  of  the  cost  of  the  farm? 

4.  A  man  desiring  to  find  the  height  of  a  tree,  places  a  12-foot  pole 
upright  54  feet  from  the  base  of  the  tree;    he  then  steps  back  6  feet,  and 
looks  over  the  top  of  the  pole  at  the  top  of  the  tree;    his  eyes  are  4 
feet  above  the  ground.     How  high  is  the  tree? 

5.  I  have,  as  the  net  proceeds  of  a  consignment  of  goods  sent  by 
me,  $3816.48,  which  the  consignor  desires  me  to  remit  by  draft  at  2  months. 
If  the  rates  of  exchange  are  f%  premium,  and  the  rate  of  interest  6%, 
what  will  be  the  face  of  the  draft? 

6.  In  a  certain  factory  are  employed  men,  women,  and  boys ;    the  boys 
receive  3  cents  per  hour,  the  women  4  cents,  and  the  men  6  cents ;    the 
boys  work  8  hours  per  day,  the  women  9  hours,  the  men  12  hours ;    the 
boys  receive  $5  as  often  as  the  women  receive  $10,  and  for  every  $10  paid 
to  the  women,  $24  are  paid  to  the  men.     How  many  are  there  of  each,  the 
whole  number  being  59? 

7.  Chicago  is  87°     35'  west.    What  is  the  standard  time  at  Chicago 
when  it  is  1  P.  M.  at  Greenwich? 

8.  From  the  middle  of  the  side  of  a  square  10-acre  field,   I  run  a 
line  cutting  off  3$-  acres.     Find  the  length  of  the  line. 

Ohio  State  List,  June,  1899. 

1.  How  would  you  present  to  a  class  the  subject  of  addition  of  frac- 
tions ?    Take  as  an  illustrative  example,  f  +  f  +  TV 

2.  A  reservoir  is  1.50  meters  wide,  2.80  meters  long,  and  1.25  meters 
deep.     Find  how  many  liters  it  contains  when  full,  and  to  what  height 
it  would  be  necessary  to  raise  it  that  it  might  contain  10  cu.  meters. 

3.  Reduce     (a)     .4685    T.   to  integers  of  lower  denominations,   and 
(b)    1.69408  to  a  common  fraction  in  its  lowest  terms. 

4.  The  boundaries  of  a  square  and  circle  are  each  40  feet.     Which  has 
the  greater  area  and  how  much? 

5.  Find  the  date  of  a  note  of  $760,  at  8%  simple  interest,  which,  when 
it  matured  December  1,  1891,  amounted  to  $919.60. 

6.  A  gentleman  wishes  to  invest  in  4J%  bonds,  selling  at  102,  so  as 
to  provide  for  a  permanent  income  of  $1620.     How  much  should  he  invest  ? 

7.  From  one-tenth  take  one-thousandth ;    multiply  the   remainder  by 
10000 ;    divide  the  product  by  one  million,  and  write  the  answer  in  words. 

8.  Bought  50  gross  of  buttons  for  25,  10,  and  5%  off,  and  disposed 
of  the  lot  for  $35.91  at  a  profit  of  12%.     What  was  the  list  price  of  the 
buttons  per  gross? 

9.  Had  an  article  cost  10%  less,  the  number  of  per  cent  gain  would 
have  been  15  more.     What  was  the  per  cent  gain?     Give  analysis. 

10.  If  the  volume  of  two  spheres  be  100  cu.  in.  and  1000  cu.  in.  re- 
spectively.    Find  the  ratio  of  their  diameters  to  the  nearest  thousandth 
of  an  inch. 

Ohio  State  List,  December,  1891. 


MENSURATION.  361 

PROBLEMS. 


1.  What  is  the  area  of  a  field  in   the   form  of   a    parallelogram,  whose 
length  is  160  rods  and  width  75  rods?  Ans.  75  A. 

2.  Find   the   area  of  a  triangle  whose  base  is  72  rods  and   altitude  16 
rods.  Ans.  3  A.  2  R.  16  P. 

3.  Two  trees  whose  heights  are  40  and  80  feet  respectively,  stand  on  op- 
posite sides  of  a  stream  30  ft.  wide.    How  far  does  a  squirrel  leap  in  jumping 
from  the  top  of  the  higher  to  the  top  of  the  lower?  Ans.  50  feet. 

4.  How   many  steps  of  3  feet  each  does  a  man  take  in  crossing  diagonal 
ly,  a  square  field  that  contains  20  acres?  Ans.  440  steps. 

5.  Find  the  cost  of  paving  a  court  150  feet  square;  a  walk  10  feet  around 
the  whole  being  paved  with   flagstones  at  54  cents  a  squaie  yard  and  the 
rest  at  31>£  cents  a  square  yard?  Ans.  $939.40. 

6.  What  is  the  area  of  a  triangle,  the  three  sides  of  which   are   respect- 
ively 180  feet,  150  feet,  and  80  feet?  Ans   5935.85  sq.  ft. 

7.  What  is  the  area  of  a  trapezium,  the  diagonal  of'which  is  110  feet,  and 
the  perpendiculars  to  the  diagonal  are  40  feet  and  60  feet  respectively  ? 

Ans.  5500  sq.  ft. 

8.  At  30  cents  a  bushel,  find   the    cost   of  a  box  of  oats,  the  box  being  & 
feet  long,  4  feet  wide  and  4  feet  deep.  Ans.  $30.85)^. 

9.  Two  trees   stand   on   opposite   sides  of  a  stream  40  feet  wide.     The 
height  of  one  tree  is  to  the  width  of  the  stream  as  8  is  to  4,  and  the  width  of 
the  stream  is  to  the   height  of  the  other  as  4  is  to  5.      What  is  the  distance 
between  their  tops  ?  Ans.  50  feet. 

10.  How  many  miles  of  furrow  15  in.,  wide,  is  turned  in  plowing  a  rect- 
angular field  whose  width  is  30  rods  and  length  10  rods  less  than  its  diagoal  ? 

Ans.  4Q%  mi. 

11.  The  sides  of  a  certain  trapezium  measure  10,  12,  14,  and  16  rods 
respectively,  and  the  diagonal,  which  forms  a   triangle  with  the  first  two 
sides,  is  18  rods;  what  is  the  area?  Ans.  163.796  sq.  rds. 

12.  Three  circles,  each  40  rods  in  diameter,  touch  each  other  externally; 
what  is  the  area  of  the  space  inclosed  between  the  circles  ? 

Ans.  64.5  sq.  rds. 

13.  How  many  square  *ncnes  in  one  face  of  a  cube  which  contains  2571353 
cubic  inches?  Ans.  18769  sq.  in. 

14.  Four  ladies  bought  a  ball  of  thread  3  inches  in  diameter;  what  por- 
tion of  the  diameter  must  each  wind  off  to  heve  equal  shares  of  the  thread? 

First,  .2743191  in. 
\  Second,  .3445792  in. 
.  -jThird?  .4912292  in. 
1  Fourth,  1.8898815  in. 

15.  A  gentleman  proposed  to  plant  a  vineyard  of  10  A.     If  he  pla'ces  the 
vines  6  feet  apart;  how  many  more  can   he  plant  by  setting  them  in  the 
quincunx  order  than  in  the  square  order,  allowing  the  plat  to  lie  in  the  form 
of  a  square,  and  no  vine  to  be  set  nearer  its  edge  than  1  foot  in  either  case? 

Ans.  1870. 

16.  Find  the  volume  generated  by        the  revolution  of   a  circle  about  a 
tangent.  Ans.  2nzRz. 

17.  How  many  feet  in  aboard  14  feet  long  and  16  inches  wide   at  one 
end  and  10  inches  at  the  other,  and  3  inches  thick?  Ans.  45^  feet. 

18.  If  I  saw  through^  of  the  diameter  of  a  round  log,  what  portion  of 
the  cut  is  made?  Ans.  .196. 


362  FINKEL'S   SOLUTION   BOOK. 

T9.    .What  is  the  surface  of  the  largest  cube  that  can  be  cut  from  a  sphere 
which  contains  14137.2  cu.  ft.?  Ans.  1800  sq.  ft. 

20.  Two  boys  are  flying  a  kite.     The  string  is  720  feet  long.      One  boy 
who  stood  directly  under  the  kite,  was  50  feet  from  the  other  boy  who  held 
the  string;  how  high  was  the  kite?  Ans.  717.8-|-feet. 

21.  How  many  pounds  of  wheat  in  a  cylindrical  sack  whose   diameter  is 
1>£  feet,  and  whose  length  is  1%  yards?     (?r:=3.1416)  s±ns.  447.31  Ib. 

22.  How  large  a  square  can  be  cut  from  a  circle  50  inches  in  diameter? 

A ns.  35.3553391  in. 

23.  How  many  bbl.  in  a  tank  in  the  form  of   the  frustum   of  a  pyramid, 
5  feet  deep,  10  feet  square  at  the  bottom  and  9  feet  square  at  the  top? 

Ans.  107.26  bbl. 

24.  From  a  circular  farm  of  270  acres,  a  father  gives  to  his   sons   equal 
circular  farms,  touching  each  other  and    the  boundary  of  the  farm.      He 
takes  for  himself  a   circular  portion   in   the  center,  equal  in  area  to  a  son's 
part,  and  reserves  the  vacant  tracts  around  his   part  for  pasture  lands  and 
gives  each  son  one  of  the  equal  spaces  left  along   the   boundary.     Required 
the  number  of  sons  and  the  amount  of  pasture  land  each  has. 

Ans.  6  sons;  8.46079  A. 

25.  At  each  angle  of  a  triangle    being  on  a  level  plain   and  having  sides 
respectively  40,  50,  and  60  feet,  stands  a  tower  whose  height  equals  the  sum 
of  the  two  sides  including  the  angle.     Required  the  length  of   a   ladder  to 
reach  the  top  of  each  tower  without  moving  at  the  base. 

Ans.  116.680316-fft. 

26.  If  the  door  of  a  room  is  4  feet  wide,  and  is  opened  to  the  angle  of  90 
degrees,  through  what  distance  has  the  outer  edge  of  the  door  passed? 

Ans.  6.2832  feet. 

27.  A  tinner  makes  two  similar  rectangular  oil  cans  whose  inside  dimen- 
sions are  as  3,  7,  and  11.      The  first  hold  8  gallons  and  the  second   being 
larger  requires  4  times  as  much  tin  as  the  other.     What  are  the  dimensions 
of  the  smaller  and  the  contents  of  the  larger  ? 

.        j  Dimensions  of  smaller  6,  14,  and  22  inches. 
1ls'    (  Capacity  of  larger  64  gallons. 

28.  An  8-inch  globe  is  covered  with  gilt  at  8  cents  per  square  inch;    find 
the  cost.  A  ns.  $16.08. 

29.  A  hollow   cylinder  6  feet   long,  whose   inner   diameter  is  1  inch  and 
outer  diameter  two  inches,  is  transformed  into  a  hollow  sphere  whose  outer 
diameter  is  twice  its  inner  diameter;  find  outer  diameter.        Ans.  3.59  in 

30.  A  circular  field  is  360  rods  in  circumference;  what  is  the  diagonal  of 
a  square  field  containing  the  same  area?  Ans.  20.3  rods. 

31.  What  is  the  volume  of  a  cylinder,  whose  length  is  9  feet  and  the  cir- 
cumference of  whose  base  is  6  feet?  Ans.  25.78  cu.  ft. 

32.  How  many  acres  in  a  square  field,  the  diagonal  being  80  rods? 

Ans.  20  acres. 

33.  How   many    cubical   blocks,  each  edge  of  which  is  ^3  of  a  foot,  will 
fill  a  box  8  feet  long,  4  feet  wide,  and  2  feet  thick.  Ans.  1728  blocks. 

34.  From  one  corner  of  a  rectangular  pyramid  6  by  8  feet,  it  is  19  feet 
to  the  apex;  find  the  dimentions  of  a  rectangular  solid  whose  dimensions  are 
as  2,  3,  and  4,  that  may  be  equivalent  in  volume.         Ans.  4,  9,  and  8  feet. 

35.*  A  solid  metal  ball,  4  inches  radius,  weighs  8  Ibs.;  what  is  the  thick- 
ness of  spherical  shell  of  the  same  metal  weighing  7%  Ib.,  the  external  di- 
ameter of  which  is  10  inches?  Ans.  1  inch. 

36.     What  is  the  difference  between  25  feet  square  and  25  square  feet? 

Ans.  600  sq.ft. 


MENSURATION.  363 

37.*  Find  the  greatest  number  of  trees  that  can  be  planted  on  a  lot  il 
rods  square,  no  two  trees  being  nearer  each  other  than  one  rod? 

Ans.  152  trees. 

38.*  A  straight  line  200  feet  long,  drawn  from  one  point  in  the  outer 
edge  of  a  circular  race  track  to  another  point  in  the  same,  just  touches  the 
inner  edge  of  the  track.  Find  the  area  of  the  track  and  its  width. 

Ans.  Area,  7r«2^=10000~  sq.  ft.;  width,  indeterminate. 

39.  The  perimeter  of  a  certain  field  in  the  form  of  an  equilateral  triangle 
is  360  rods;  what  is  the  area  of  the  field  ?  Ans.  543.552  sq.  rd. 

40.  A  room  is  18  feet  long,  16  feet  wide,  and  10  feet  high.      What  length 
of  rope  will  reach  from  one  upper  corner  to  the  opposite  upper  corner  and 
touch  the  floor?  Ans.  35  3  ft. 

41.  How  many  bushels  of  wheat  in  a  box  whose  length  is  twice  its  width, 
and  whose  width  is  4  times  its  height;  diagonal  being  9  feet? 

Ans.  25  bu.,  nearly. 

42  Find  the  area  of  a  circular  ring  whose  breadth  is  2  inches  and  inside 
diameter  9  inches.  Ans.  69.1152  sq.  in. 

43  *     A  round  stick  of  timber  12  feet  long,  8  inches  in  diameter  at  one 
end  and  16  inches  at  the  other,  is  rolled  along  till  the  larger  end  describes  a 
complete  circle.      Required  the   circumference   of  the  circle. 

Ans.  150.83  feet. 

44.  A  fly  traveled  by  the  shortest  possible  -route  from  the  lower  corner 
to  the  opposite  upper  corner  of  a  room  18  feet  long,  12  feet  wide  and  10  feet 
high.  Find  the  distance  it  traveled  Ans.  28.42534  feet. 

45.*  From  the  middle  of  one  side  and  through  the  axis  perpendicularly 
of  a  right  triangular  prism,  sides  12  inches,  I  cut  a  hole  4  inches  square. 
Find  the  volume  removed.  Ans.  138.564064  cu.  in. 

46.*  Two  isosceles  triangles  have  equal  areas  and  perimeters.  The  base 
of  one  is  24  feet,  and  one  of  the  equal  sides  of  the  other  is  29  feet.  The 
area  of  both  is  10  times  the  area  of  a  triangle  whose  sides  are  13,  14,  and  15 
feet.  Find  the  perimeters  and  altitudes. 

Ans.  Perimeters,  98  feet;    altitudes  35  and  21  feet. 

47.  A  grocer  at  one  straight  cut  took  off  a  segment  of  a  cheese  which 
had  ^  of  the  circumference,  and  weighed  3  pounds;  what  did  the  whole 
weigh?  A ns.  33.023  lb. 

48.*  A  twelve  inch  ball  is  in  a  corner  where  walls  and  floor  are  at  right 
angles;  what  must  be  the  diameter  of  another  ball  which  can  touch  that 
ball  while  both  touch  the  same  floor  and  the  same  walls? 

Ans.  3.2154  in.  or  44.7846  in. 

49.  What  will  it  cost  to  paint  a  church   steeple,  the  base  of  which  is  an 
octagon,  6  feet  on  each  side,  and  whose  slant   height  is  80  feet,  at  30  cents 
per  square  yard?  Ans.  $64. 

50.  A  tree  48  feet  high  breaks  off;  the  top  strikes  the  level  ground  24  feet 
from  the  bottom  of  the  tree;  find  the  height  of  the  stump.        Ans.  18  feet. 

51.  How  many  acres  in  a  square  field  whose  diagonal  is  5^  rods  longer 
than  one  of  its  sides?  .  Ans.  160.6446  sq.  rd. 

52.*  Three  poles  of  equal  length  are  erected  on  a  plane  so  that  their  tops 
meet,  while  their  bases  are  90  feet  apart,  and  distance  from  the  point  where 
the  poles  meet  to  the  center  of  the  triangle  below  is  65  feet.  What  is  the 
length  of  the  poles  ?  Ans.  83.23  feet. 

53.  A  field  contains  200  acres  and  is  5  times  as  iong  as  wide.  What  will 
it  cost  to  fence  it,  at  a  dollar  per  rod  ?  Ans.  $960. 

54.*  What  is  the  greatest  number  of  plants  that  can  be  set  on  a  circular 
piece  of  ground  100  feet  in  diameter,  no  two  plants  to  be  nearer  each  other 
than  2  feet  and  none  nearer  the  circumference  than  1  foot?  Ans.  2173. 


364  FINKEL'S   SOLUTION  BOOK. 

55.  The  axes  of  an  ellipse  are  100  inches  and  60  inches;   what  is  the  dif- 
ference in  area  between  the  ellipse  and  a  circle  having  a  diameter  equal  to 
the  conjugate  axis  ?  A  ns.  600  ^=1884.96  sq.  in. 

56.  Find  the  diameter  of  a  circle  of  which  the  altitude  of  its  greatest  in- 
scribed triangle  is  25  feet.  Ans.  33  J/  feet. 

57.  If  we  cut  from  a  cubical  block  enough  to  make  each   dimension  1 
inch  shorter,  it  will  lose  1657  cubic  inches,  what  are  the  dimensions? 

58.  Show  that  the  area  of  a  rhombus  is  one-half  the  rectangle  formed  by 
its  diagonals.     Noble  Co.  Ex.  Test. 

59.  The   length    and   breadth  of  a  rectangular   field  are  in  the  ratio  of  4 
to  3.     How  many  acres  in  the  field,  if  the  diagonal  is  100  rods  ? 

60.  A  spherical  vessel  30  inches  in  diameter  contains  in  depth,  1  foot  of 
water;  how  many  gallons  will  it  take  to  fill  it?     Holmes  Co.  Ex.  Test. 

Ans.  39  gallons. 

61.  A  field  is  40  rods  by  80  rods.      How   long  a  line  from  the  middle  of 
one  end  will  cut  off  7)£  acres?  Ans.  80.6  rd.,  nearly. 

62.  A  ladder  20  feet  long  leans  against  a  perpendicular  wall  at  an  angle 
of  30°.     How  far  is  its  middle  point  from  the  bottom  of  the  wall? 

Ans.  10  feet. 

63.  Four  towers,  A  125  feet  high,  B  75  feet,  C  160  feet,  and  D  65  feet, 
stand  on  the  same   plane.      B  due  south  and  40  rods  from  A;  C  east  of  B 
and  D  south  of  C.    The  distance  from  A  to  C  plus  the  distance  from  C  to  B 
is  half  a  mile,  and  the  distance  from  D  to  B  is  82}^  yd.  farther  than  the  dis- 
tance from  C  to  D.     What  length  of  line  is  required  to  connect  the  tops  of 
AandD?  Ans.  240+rus. 

64.  Find  the  volume  of  the  largest  square  pyramid  that   can  be  cut  from 
a  cone  9  feet  in  diameter  and  20  feet  high?  Ans.  270  cu.  ft. 

65.  A  rectangular  lawn  60yd.  long  and  40  yd.  wide  has  a  walk  6ft.  wide 
around  it  and  paths  of  the  same  width  through  it,  joining  the    points  of  the 
opposite  sides.     Find  in  square  yards  the  area  of  one  of  the   four  plats  in- 
closed by  paths.  Ans.  459  sq.  yd. 

66.  Which  has  the  greater  surface,  a  cube  whose  volume  is  13.824  cu.  ft., 
or  a  rectangular  solid  of  equal  volume  whose  length  is  twice  its   width,  and 
its  width  twice  its  height?  Ans.  Rect.  576  sq.  ft,  more. 

67.  The  volume  of  a  rectangular  tin  can  is  3  cu.  ft.  1053  cu.  in.;  its  di- 
mensions are  in  the  proportion  of  11,  7,  and  3.      Find  the  area  of  tin  in  the 
can.  Ans.  16%  sq.  ft. 

68.  A   conical   well    has   a  bottom  diameter  of  28  ft.  3  in.,  top  diameter 
56  ft.  6  in.,  and  depth  23  ft.  1.2  in.     Find  its  capacity  in  barrels. 

Ans.  8023  bbl. 

69.  A  cylindrical  vessel  1  foot  deep  and  8  inches  in  diameter  was  |f  full 
of  water;  after  a  ball  was  dropped  into  the  vessel  it  was  full.      Find  the  di- 
ameter of  the  ball.  Ans.  6  inches. 

70.  Two  logs  whose  diameters  are  6  feet  lie  side  by  side.   What  is  the  di- 
ameter of  a  third  log  placed  in  the  crevice  on  top  of  the  two,  if  the  pile  is  9 
feet  high?.  Ans.  4  ft. 

71.  Circles  6  and  10  feet  in  diameter  touch  each  other;  if  perpendiculars 
from  the  center  are  let  fall  to  the  line  tangent  to  both  circles,  how  far  apart 
will  they  be?  Ans.  7.756  ft. 

72.  What  are  the  linear  dimensions  of  a  rectangular  box  whose  capacity 
is  65910  cubic  feet;  the  length,  breadth,  and  depth  being  to  each  other  as 
5,  3,  and  2?  Ans.  65,  39,  and  26  ft. 

73.  The  perimeter  of  a  piece  of  land  in  the  form  of  an  equilateral  trian- 
gle is  624  rods;  what  is  the  area?  Ans.  117  A.  13  31  P. 


MENSURATION.  365 

74.  Four  logs  4  feet  in  diameter  lay  side  by  side   and   touch  each  other; 
on  these  and  in  the  crevices  lay  three  logs  3  feet  in  diameter;  on  these  three 
and  in  the  crevices   lay   two   logs   2  feet   in  diameter;    what  is  the  diame- 
ter of  a  log  that  will  lay  on  the  top  of  the  pile  touching  each  of  the  logs  2 
feet  in  diameter  and  the  middle  one  of  the  logs  3  feet  in  diameter? 

Ans. 

75.  What  will  it  cost  to  gild  a  segment  of   a  sphere  whose  diameter  is  6 
inches;  the  altitude  of  the  segment  being  2  inches,  at  5  ^  per  square  inch? 

76.  A  grocer  cut  off  the  segment  of  a  cheese,  and  found  it  took  £  of  the 
circumference.     What  is  the  weight  of  the  whole  cheese,  if  the  segment 
weighed  \%  Ibs  ?  Ans.  52.0228+lbs. 

77.  Two  ladders  are  standing  in  the  street  20  feet  apart.       They   are  in- 
clined equally  toward  each  other  at  the  top,  forming  an  angle  of  45°.     Find, 
by  arithmetic,  the  length  of  the  ladders?  Ans.  26.13  ft. 

Union  Co.  Ex.  List. 

78.  Two  trees  stand  on  opposite  sides  of  a  stream  120  feet  wide;  the  height 
of  one  tree  is  to  the  width  of  the  stream  as  5  is  to  4,  and   the    width  of  the 
stream  is  to  the  height  of  the  other  as  5  is  to  4;  what  is  the  distance  between 
their  tops?  Ans.  131.58— ft. 

79.  How  many  gallons  of  water  will  fill  a  circular  cistern  6  feet   deep 
and  4  feet  in  diameter?  Ans.  564.0162  gal. 

80.  A  cube  of   silver,   whose   diagonal  is  6   inches,   was   evenly    plated 
with  gold;  if  4  cubic  inches  of  gold  were  used,  how  thick  was  the  plating? 

Ans.  iV  in. 

81.  Required  the  distance  between  the   lower  corner   and   the   opposite 
upper  corner  of  a  room  60  feet  long,  32  feet  wide,  and  51  feet  high  ? 

Ans   85  ft. 

82.  How  deep  must  be  a  rectangular  box   whose  base   inside  is   4  inches 
by  4  inches  to  hold  a  quart,  dry  measure?  Ans.  4.2  cu  in 

83.  A   fly   is  in  the  center  of  the  floor  of  a  room  30  feet  long,  20  feet 
wide,  and  12  feet  high.     How  far  will  it  travel  by  the   shortest   path  to  one 
of  the  upper  corners  of  the  ceiling?  Ans.  v/709-j-ft 

84.  A  corn  crib  25  feet  long  holds  125  bushels.     How  many  bushels  will 
one  of  like  shape  and  35  feet  long  hold  ? 

85.  Let  a  cube  be  inscribed  in  a  sphere,  a  second  sphere  in  this  cube,  a 
second  cube  in  this  sphere,  and  so  on;  find  the  diameter  of  the  7th  sphere, 
if  thatof  the  first  is  27  inches.     (2).  What  is  the  volume  of  all  the  spheres 
so  inscribed  including  the  first?  Ans. . 

86.  The  area  of  a  rectangular  building  lot  is  720   sq.   ft  ;  its  sides  are  as 
4  to  5;  what  will  it  cost  to  excavate  the  earth  7  feet  deep   at   36^  per  cubic 
yard?  Ans.  $67.20. 

87.  A  owns  ^    and  B  the  remainder  of  a  field   60  rods   long  and  30  rods 
wide  at  one  end  and  20  rods  wide  at  the  other  end,  both  ends  being  parallel 
to  the  same  side  of  the  neid      They  propose  to  lay  out  through  it,  parallel 
with  the  ends,  a  road  one  rod  wide  leaving  A's  %  of  the   remainder  at  the 
wide  end  and  B's  %  at  the  narrow  end  of  the  field.     Required  the  location 
and  area  of  the  road.  Ans . 

88.  The  diameter  of  a  circular  field  is  240  rods.      How   much   grass  will 
be  left  after  7  horses  have  eaten  all  they  can  reach,   the  ropes  which  are  al- 
lowed them  being  of  equal  lengths  and  attached  to  posts  so  located  that  each 
can  touch  his  neighbor's  territory  and  none  can  reach  beyond  the  boundary 
of  the  field?  Ans.  62.831853  A. 

89.  What  is  the  diameter  of  a  circle  inclosing  three  equal  tangent  circles, 
if  the  area  inclosed  by  the  three  equal  circles  is  ]  acre?         Ans. . 


366  FINKEIv'S  SOLUTION   BOOK. 

90.  What  is  the  diameter  of  a  circle  inclosing  four    equal  tangent  circles 
each  being  tangent  to  the  the  required    circle,    if   the    area   inclosed  by  the 
four  equal  circles  is  I  acre?  Ans.  /?=4j/[5(4 — 7r)](v/2+l)-f-(4 — TT). 

91.  What  is  the  greatest  number  of  stakes  that   can   be   driven  one  foot 
apart  on  a  rectangular  lot  whose  length  is  30  feet  and  width  20  feet? 

A  ns. . 

92.  What  is  the  greatest  number  of  inch  balls  that  can  be  put  in  a  box  15 
inches  long,  9  inches  wide,  and  6  inches  high?  Ans. . 

93.  A  conical  vessel  6  inches  in  diameter  and  10  inches    deep  is  full  of 
water.     A  heavy  ball  8  inches  in  diameter,  is  put  into  the  vessel;  how  much 
water  will  flow  out?  Ans . 

94.  How  far  above  the  surface  of  the  earth  would  a  person  have  to  ascend 
in  order  that  %  of  its  surface  would  be  visible?  Ans.  8000  mi. 

95.  Where  must  a  frustum  of   a   cone   be   sawed    in    two   parts,  to  have 
equal  solidities,  if  the  frustum  is  10  feet  long,  2  feet  in  diameter  at  one  end, 
and  6  feet  at  the  other?  Ans  . 

96.  At  the  three  corners  of  a  rectangular  field  50  feet   long   and   40  feet 
wide,  stands  three  trees  whose   heights   are  60,  80,  and  70  feet.      Locate 
the  point  where  a  ladder  must  be  placed  so   that  without   moving  it  at  the 
base  it  will  touch  the  tops  of  the  three  trees,  and  find   the  length  of  the  lad- 
der.    What  must  be  the  height  of  a  tree  at   the  fourth  corner   so  that  the 
same  ladder  will   reach  the   top,  the  foot  of  the  ladder  not  being  moved? 

Ans. . 

97.  A  horse  is  tied  to  a  corner  of  a  barn  50  feet   long  and   30  feet  wide; 
what  is  the  area  of  the  surface  over  which  the  horse  can   graze,  if  the   rope 
is  80  feet  long?  Ans. . 

98      How  many  cubic  feet  in  a  stone  32  feet  high,   whose   lower  base  is  a 
rectangle,  10  feet  by  4  feet  and  the  upper  base  8  feet  by  1%  feet? 

Ans.  805^  cu.  ft. 

99.  To  what  height  above  the   ground   would  a   platform,    10  feet  by  6 
feet,  have  to  be  elevated  so  that  720  sq.  ft.  of  surface  would  be  invisible  to  a 
man  standing  at  the  center  of  the  platform,  the  man  being  5  feet  high?. 

100.  Required  the  side  of  the   least   equilateral   triangle   that   will  cir- 
cumscribe seven  circles,  each  20  inches  in  diameter.          Ans.  89  28203  in. 

101.  Required  the  sides  of  the  least  right  triangle  that  will  circumscribe 
seven  circles  each  20  inches  in  diameter.  Ans    123.9320  in.  and 

107.3205  in. 

102.  How  long  a  ladder  will  be  required  to  reach  a  window  40  feet  from 
the  ground,  if  the  distance  of  the  foot  of  the  ladder  from  the  wall  is  f  of  the 
length  of  the  ladder.  Ans.  50  ft. 

103.  A  circular  park  is  crossed  by  a  straight   path   cutting  off  ^   of  the 
circumference;  the  part  cut  off  contains  10  acres       Find  the  diameter  of  the 
park.  Ans.  150  rd.,  nearly 

104.  Find  the  length  of  the  minute-hand  of  a  clock,  whose  extreme  point 
moves  5  ft.  5.9736  in.,  in  1  da.  18  hr. ?  Ans.  \  in. 

105.  A,  B,  and  C,  own  a  triangular  tract  of  land.  Their  houses  are  located 
at  the  vertices  of  the  triangle;     where  must  they  locate   a  well  to  be  used  in 
common  so  that  the  distance  irom  the  houses  to  the  well   will  be  the  same, 
the  distance  from  A  to  B  being  120  rods,  from    B  to    C  90  rods  and  A  to  C 
80  rods.  Ans.  — . 

106.  A  horse  is  tethered  from  one   corner    of  an    equilateral    triangular 
building  whose  sides  are  100  feet,  by   a  rope  175  feet   long.     Over  what  area 
can  he  graze?  Ans.  90021.109181  sq.  ft. 

107.  Find  the  area  of  the  triangle  formed  by   joining   the   centers  of  the 
squares  constructed  on  the  sides  of  an    equilateral  triangle,  whose  sides  are 
20  feet  ?  A  ns. 


GEOMETRY. 


367 


GEOMETRY. 


I.     DEFINITIONS. 

i.  Geometry  is  that  branch  of  mathematics  which  deduces 
the  properties  of  figures  in  space  from  their  defining  conditions, 
by  means  of  assumed  properties  of  space.  —  Century  Dictionary. 

ri3  Plane 


j.  Platonic 
Geometry  - 


2.  Geometry 


1.  Metrical 
Geometry 


12.  Pure 

Geometry  -|  „ 

1 8 

22.  Conic  Secions 


Geometry 

Solid 

Geometry 


.38.Trigon'try- 
2i.  Analytical  Geometry 


2.  Descriptive  Geometry,  or  Projective  Geometry 


13.  Plane 

Trig. 
28.  Analytical 

Trig. 
33.  Spherical 

Trig. 


3.  Metrical  Geometry  is  that  branch  of  Geometry  which 
treats  of  the  length  of  lines  and  the  magnitudes  of  angles,  areas, 
and  solids. 

The  fundamental  operation  of  metrical  relations  is  MEASUREMENT. 
The  geometry  of  Euclid  and  the  Ancients  is  almost  entirely  metrical.  The 
theorem,  The  square  described  on  the  hypotenuse  of  a  right-angled  triangle 
is  equal  to  the  sum  of  the  squares  described  on  the  other  two  sides,  is  a  theo- 
rem of  metrical  geometry. 

4.  Descriptive   Geometry,  also  called  Projective  Geom- 
etry, Modern  Synthetic  Geometry,  and  Geometry  of  Position,  is 
that  branch  of  Geometry  which  treats  of  the  positions,  the  direc- 
tions, and  intersections  of  lines,  the  loci  of  points,  and  the  nature 
and  character  of  curves  and  surfaces. 

The  fundamental  operations  of  Descriptive  Geometry  are  PROJECTION 
and  SECTION.  Many  of  the  theorems  of  Descriptive  Geometry  are  very 
old,  dating  as  far  back  as  the  time  of  Euclid,  but  the  theories  and  methods 
which  make  of  these  theorems  a  homogeneous  and  harmoneous  whole  is 
modern  having  been  discovered  or  perfected  by  mathematicians  of  an  age 
nearer  our  own,  such  as  Monge,  Carnot,  Brianchon,  Poncelet,  Moebius, 
Steiner,  Chasles,  von  Staudt,  etc.,  whose  works  were  published  in  the 
earlier  half  of  the  present  century.  Of  the  synonymous  terms  I  have  used 
to  designate  this  geometry  of  which  I  am  speaking,  the  term,  Modern  Syn- 
thetic Geometry  is  the  most  comprehensive.  Descriptive  Geometry  was 
invented  by  Gaspard  Monge  (1746-1818)  in  1794  and  at  that  time  embraced 
only  the  theory  of  making  projections  of  any  accurately  defined  figure  such 
that  from  these  projections  can  be  deduced,  not  only  the  projective  proper- 
ties of  the  figure,  but  also  its  metrical  properties.  Now  this  term  is  used 
to  designate  the  entire  theory  and  development  of  geometry  as  embraced 
in  the  above  definition. 


368 


FINKEL'S   SOLUTION   BOOK. 


The  problem,  To  draw  a  third  straight  line  through  the  inaccessable 
point  of  intersection  of  two  (converging)  straight  lines,  is  both  metrical 
and  descriptive,  that  is  to  say,  the  required  line  may  be  found  either  by 
metrical  or  descriptive  geometry,  but  the  method  by  Descriptive  Geometry 
is  far  the  simpler. 

The  following  are  the  solutions  by  both  methods : 


METRICAL. 


I.  Given  the  two  converg- 
ing lines  AB  and  CD  which 
do  not  intersect  in  an  acces- 
sable  point. 

II.  Required    to  draw    a 
third  line  through  the  inac- 
cessable point  K. 


1.  Draw  the  transversal 
LM,  intersecting  AB  and 
CD  in  E  and  F  respec- 
tively. 

2  .     Draw  A^parallel  to  EF 

and  intersecting  AB  and 
CD  in  G  and  H  respec- 
tively. 

3  .     Divide  EF  in  any  ratio  , 

say  1:2,  and  let  Q  be  the 
point  of  division. 

4.  Divide  GH  in  the  same 
ratio   and   let  R  be  the 
point  of  division. 

5.  The  line  through  QR 
is  the  line  required. 


\J 

I 


i 


fl.  Suppose  the  line  join- 
ing the  inaccessable  point 
K  and  the  point  Q  to  in- 
tersect NP  in  R',  if  not 
inR. 


DESCRIPTIVE. 


1.  Choose  some  point  P  out- 
side the  two  given  straight 
lines  AB  and  CD. 

2.  Pass    through    this    point 
any  number  of  transversals, 
as  FPt  HP,  KP. 

3.  Draw  the    diagonals  FGr 
HE,  HI,  and  KG. 

4.  The  points  of  intersection 
L  and  M  lie  upon  the  line 
which    passes    through    the 
point  of  intersection  of  AB 
and  CD. 

The  proof  of  this  follows  from 
the  important  harmonic  prop- 
erties of  a  quadrangle. 


t 


^5r     c/  u 


GEOMETRY. 


369 


, 


Then,  from  similar  tri- 
angles, KR':KQ=R'G: 
QE. 

Also,  KR'  :  KQ=RH\ 


.-.  R'G:QE=R'H:QE. 

But,  Q£:QF=1:2.  By 

Hyph. 


9, 


=l:2.  By 

Const. 

.'.RG:RH=R'G:R'H. 

.'.RG=R'G    and    the 
point  ^'coincides  with./?. 


Many  of  the  properties  of  the  Conic  Sections  which  are  estab- 
lished with  great  labor  and  difficulty  by  Analytical  Geometry  are 
easily  and  elegantly  proved  by  Descriptive  Geometry.  Descriptive 
Geometry  stands  among  the  first  of  the  branches  of  pure  mathe- 
matics in  point  of  interest  and  simplicity  of  its  methods.  The 
best  works  on  this  subject  are  Luigi  Cremona's  Elements  of  Pro- 
jective  Geometry,  translated  by  Charles  Leudesdorf,  and  Theodore 
Reye's  Lectures  on  Geometry  of  Position,  Part  I.,  translated  by 
Thomas  F.  Holgate. 

II.     ON  GEOMETRICAL  REASONING. 

5.  On  Geometrical  Reasoning.  We  are  accustomed  to 
speak  of  mathematical  reasoning  as  being  above  all  other,  in 
accuracy  and  soundness.  This  is  not  correct,  if  we  mean  by 
reasoning  the  comparing  together  of  different  ideas  and  pro- 
ducing other  ideas  from  the  comparison ;  for,  in  this  view,  mathe- 
matical reasonings  and  all  other  reasonings  correspond  precisely. 
The  nature  of  establishing  mathematical  truths,  however,  is 
totally  different  from  that  of  establishing  a  truth  in  history, 
political  economy,  or  metaphysics,  and  the  difference  is  this,  viz., 
instead  of  showing  the  contrary  of  the  proposition  asserted  to 
be  only  improbable,  it  proves  it  at  once  to  be  absurd  and  impos- 
sible. For  example,  suppose  one  were  to  ask  for  the  proof  of 
the  assassination  of  Caesar,  what  would  be  the  method  of  proof? 
No  one  living  to-day  is  absolutely  certain  that  Caesar  was  assas- 
sinated, and,  in  order  to  establish  this  truth,  we  refer  to  the  testi- 
mony of  historians,  men  of  credit,  who  lived  and  wrote  their 
accounts  in  the  very  time  of  which  they  write;  the  statements 
of  these  historians  have  been  received  by  succeeding  ages  as 
true ;  and  succeeding  historians  have  backed  their  accounts  by  a 
mass  of  circumstantial  evidence  which  makes  it  the  most  improb- 
able thing  in  the  world  that  the  account  or  any  particular  part  of 
it  is  false.  In  this  way  we  have  proved  that  the  truth  of  the 


370  FINKEL'S   SOLUTION  BOOK. 

statement  rests  on  a  very  high  degree  of  probability,  though  it 
does  not  rise  to  absolute  certainty. 

"In  mathematics,  the  case  is  wholly  different.  It  is  true  that 
the  facts  asserted  in  these  sciences  are  of  a  nature  totally  distinct 
from  those  of  history;  so  much  so,  that  a  comparison  of  the 
evidence  of  the  two  may  almost  excite  a  smile.  But  if  it  be 
remembered  that  acute  reasoners,  in  every  branch  of  learning, 
have  acknowledged  the  use,  we  might  almost  say  the  necessity 
of  a  mathematical  education,  it  must  be  admitted  that  the  points 
of  connection  between  these  pursuits  and  others  are  worth  attend- 
ing to.  They  are  the  more  so,  because  there  is  a  mistake  into 
which  several  have  fallen,  and  have  deceived  others,  and  per- 
haps themselves,  by  clothing  some  false  reasoning  in  what  they 
called  a  mathematical  dress,  imagining  that,  by  the  application 
of  mathematical  symbols  to  their  subject  they  secured  mathe- 
matical argument.  This  could  not  have  happened  if  they  had 
possessed  a  knowledge  of  the  bounds  within  which  the  empire 
of  mathematics  is  contained.  That  empire  is  sufficiently  wide, 
and  might  have  been  better  known,  had  the  time  which  has  been 
wasted  in  aggressions  upon  the  domains  of  others,  been  spent 
in  exploring  the  immense  tracts  which  are  yet  untrodden/'*  In 
establishing  a  mathematical  truth,  instead  of  referring  to  authority, 
we  continually  refer  our  statements  to  more  and  more  evident 
statements,  until  at  last  we  come  either  to  definitions  or  to  state- 
ments so  evidently  true,  that  to  deny  them  would  prove  the  un- 
soundness  of  him  who  makes  the  denial. 

Geometry  must  have  recourse  to  the  outside  world  for  its 
first  notions  and  premises,  and  is,  therefore,  a  natural  science. 

Yet  there  is  a  great  difference,  between  it  and  the  other  natural 
sciences.  For  example,  contrast  Geometry  and  Chemistry.  Both 
derive  their  constructive  materials  from  sense-perception ;  but 
while  Geometry  is  compelled  to  draw  only  its  first  results  from 
observation  and  is  then  in  a  position  to  move  forward  deductively 
to  other  results  without  being  under  the  necessity  of  making 
fresh  observations,  Chemistry,  on  the  other  hand,  is  still  com- 
pelled to  make  observations  and  to  have  recourse  to  nature. 


III.  ON  THE  ADVANTAGES  DERIVED  FROM  THE 
STUDY  OF  GEOMETRY,  AND  MATHEMATICS 
IN  GENERAL. 

6.  On  the  Advantages  derived  from  the  Study  of 
Geometry  and  Mathematics  in  General.  The  story 
is  told  of  Abraham  Lincoln  that  before  he  began  the  study  of 
law,  he  worked  through  Euclid  in  order  to  give  his  mind  that 
training  in  logical  thinking  so  necessary  to  a  successful  lawyer; 

*DeMorgan,  Study  of  Mathematics. 


GEOMETRY.  371 

and  his  great  success  as  a  lawyer  and  statesman  is  largely  to  be 
attributed  to  the  discipline  he  thus  received. 

There  should  be  no  conflict  between  the  sciences  and  the 
classics.  A  student  taking  a  college  course  should  give  his  time 
to  study  in  both.  The  study  of  language  enables  a  person  to 
express  his  thoughts  accurately  and  clearly  while  the  study  of 
the  sciences  provides  him  with  thoughts  worthy  of  expression. 
How  far  each  of  these  two  great  departments  should  be  pursued 
by  the  student,  must  be  determined  by  the  student  himself.  But 
certainly  neither  should  be  pursued  exclusively.  Yet  if  one 
were  to  pursue  one  or  the  other  of  these  two  great  departments 
of  knowledge  exclusively,  I  heartily  agree  with  Professor  Earnst 
Mach  who  says,  "Here  I  may  count  upon  assent  when  I  say 
that  mathematics  and  the  natural  sciences  pursued  alone  as 
means  of  instruction  yield  a  richer  education  in  matter  and 
form  a  more  general  education,  an  education  better  adapted  to 
the  needs  and  spirit  of  the  time,  than  the  philological 
branches  pursued  alone  would  yield."*  As  to  mathematics, 
"It  is  admitted  by  all  that  a  finished  or  even  a  competent 
reasoner  is  not  the  work  of  nature  alone ;  the  experience  of  every 
day  makes  it  evident  that  education  develops  faculties  which 
would  otherwise  never  have  manifested  their  existence.  It  is, 
therefore,  as  necessary  to  learn  to  reason  before  we  can  expect 
to  be  able  to  reason,  as  it  is  to  learn  to  swim  or  fence,  in  order 
to  attain  either  of  these  arts.  Now,  something  must'be  reasoned 
upon,  it  matters  not  much  what  it  is,  provided  it  can  be  reasoned 
upon  with  certainty.  The  properties  of  mind  or  matter,  or  the 
study  of  languages,  mathematics,  or  natural  history,  may  be 
chosen  for  this  purpose.  Now,  of  all  these,  it  is  desirable  to 
choose  the  one  which  admits  of  the  reasoning  being  verified,  that 
is,  in  which  we  can  find  out  by  other  means,  such  as  measurement 
and  ocular  demonstrations  of  all  sorts,  whether  the  results  are 
true  or  not.  .  .  .  Now  the  mathematics  are  peculiarly  well 
adapted  for  this  purpose,  on  the  following  grounds : 

i°.  Every  term  is  distinctly  explained,  and  has  but  one  mean- 
ing, and  it  is  rarely  that  two  words  are  employed  to  mean  the 
same  thing. 

2°.  The  first  principles  are  self-evident,  and,  though  derived 
from  observation,  do  not  require  more  of  it  than  has  been  made 
by  children  in  general. 

3°.  The  demonstration  is  strictly  logical,  taking  nothing  for 
granted  except  the  self-evident  first  principles,  resting  nothing 
upon  probability,  and  entirely  independent  of  authority  or 
opinion. 

4°.  When  the  conclusion  is  attained  by  reasoning,  its  truth 
or  falsehood  can  be  ascertained,  in  -geometry  by  actual  measure- 

*See  Professor  Mach's  Popular  Scientific  Lectures,  "On  Instruction  in  the  Classics 
and  Sciences."  Also  Grant  Allen's  Article  in  the  Oct.  No.  of  the  Cosmopolitan  for  1897. 


372  FINKEL'S  SOLUTION  BOOK. 

ment,  in  algebra  by  common  arithmatical  calculation.  This  gives 
confidence,  and  is  absolutely  necessary,  if,  as  was  said  before, 
reason  is  not  to  be  instructor,  but  pupil. 

5°.  There  are  no  words  whose  meanings  are  so  much  alike 
that  the  ideas  which  they  stand  for  may  be  confounded.  Be- 
tween the  meanings  of  terms  there  is  no  distinction,  except  ab- 
solute distinction,  and  all  adjectives  and  adverbs  expressing  dif- 
ference of  degree  are  avoided.  Thus  it  may  be  necessary  to  say, 
"A  is  greater  than  B ;"  but  it  is  entirely  unimportant  whether 
A  is  very  little  greater  than  B  or  very  much  greater  than  B. 
Any  proposition  which  includes  the  foregoing  assertion  will 
prove  its  conclusions  generally,  that  is,  for  all  cases  in  which  A 
is  greater  than  B,  whether  the  difference  be  great  or  little.  .  .  . 

These  are  the  principal  grounds  on  which,  in  our  opinion,  the 
utility  of  mathematical  studies  may  be  shown  to  rest,  as  a  dis- 
cipline for  the  reasoning  powers.  But  the  habits  of  mind  which 
these  studies  have  a  tendency  to  form  are  valuable  in  the  highest 
degree.  The  most  important  of  all  is  the  power  of  concen- 
trating the  ideas  which  a  successful  study  of  them  increases 
where  it  did  exist  and  creates  where  it  did  not.  A  difficult 
position,  or  a  new  method  of  passing  from  one  proposition  to 
another,  arrests  all  the  attention  and  forces  the  united  faculties 
to  use  their  utmost  exertions.  The  habit  of  mind  thus  formed 
soon  extends  itself  to  other  pursuits,  and  is  beneficially  felt  in 
all  the  business  of  life. 

"As  a  key  to  the  attainment  of  other  sciences,  the  use  of  the 
mathematics  is  too  well  known  to  make  it  necessary  that  we 
should  dwell  on  this  topic.  In  fact,  there  is  not  in  this  country 
any  disposition  to  undervalue  them  as  regards  the  utility  of  their 
applications.  But  though  they  are  now  generally  considered  as 
a  part,  and  a  necessary  one,  of  a  liberal  education,  the  views 
which  are  still  taken  of  them  as  a  part  of  education  by  a  large 
proportion  of  the  community  are  still  very  confined."* 

The  advantages  derived  from  a  study  of  geometry,  though 
very  great,  are  only  part  of  those  to  be  derived  from  a  thorough 
course  of  study  in  mathematics.  The  eminent  mathematician 
Cayley,  "the  central  luminary,  the  Darwin  of  the  English  School 
of  Mathematicians,"  as  Sylvester  calls  him,  said  once  that  if 
he  had  to  make  a  defence  of  mathematics  he  would  do  it  in  the 
manner  in  which  Socrates,  in  Plato's  "Republic"  defended  jus- 
tice. Justice,  according  to  the  Greek  sage,  was  a  thing  desir- 
able, in  itself  and  for  its  own  sake,  quite  irrespective  of  the 
worldly  advantages  which  might  accompany  a  life  of  virtue 
and  justice.  So  just  for  the  sake  of  learning  the  beauties  and 
the  purest  truths  which  mathematics,  the  oldest  and  the  noblest, 
the  grandest  and  the  most  profound  of  all  sciences,  represents, 

*DeMorgan,  The  Study  of  Mathematics. 


GEOMETRY.  373 

would  it  be  worth  while  to  make  ourselves  acquainted  with  its 
uses  as  an  educational  medium  and  the  application  it  finds  in 
other  sciences?  Sylvester  says,  "The  world  of  ideas  which 
mathematics  discloses  or  illuminates,  the  contemplation  of  divine 
beauty  and  order  which  it  induces,  the  harmonious  connection 
of  its  parts,  the  infinite  hierarchy  and  absolute  evidence  of  truths 
with  which  mathematical  science  is  concerned,  these,  and  such 
like,  are  the  surest  grounds  of  its  title  to  human  regard." 
Sylvester,  twenty-five  years  ago  called  the  attention  of  the  Royal 
Society  to  the  parallelism  between  the  mathematical  and  musical 
ethos :  music  being  the  mathematics  of  the  senses,  mathematics 
the  music  of  reason;  the  soul  of  each  the  same.  Music  the 
dream,  mathematics  the  working  life ;  each  to  receive  its  con- 
summation from  the  other,  when  the  human  intelligence  elevated 
to  its  perfect  type,  shall  shine  forth  glorified  in  some  future 
Beethoven-Gauss. 

There  is  surely  something  in  the  beauty  of  the  truths  them- 
selves. They  enrich  us  by  our  mere  contemplation  of  them. 
What  a  charm  and  what  a  wealth  of  delight  and  self-content- 
ment does  the  finding  of  mathematical  truths  afford.  In  this 
science,  of  which  geometry  is  one,  out  of  a  few  postulates  and 
germinating  truths,  the  mind  of  man  can  gradually  unfold  a 
system  of  new  and  beautiful  truths  never  dreamt  of  before. 
Locke  says,  "The  mathematician  from  very  plain  and  easy  be- 
ginnings, by  gentle  degrees,  and  a  continued  chain  of  reason- 
ings, proceed  to  the  discovery  and  demonstration  of  truths  that 
appear  at  first  sight  beyond  human  capacity."  Because  mathe- 
matics is  a  science  of  pure  reason  and  rigorous  logic  a  mathe- 
matician may  forget  all  the  preceding  propositions  of  his  science 
and  still  be  able  to  guide  himself  with  the  utmost  confidence 
through  the  labyrinth  of  ideas  and  reach  its  exit,  if  he  only 
keeps  clearly  before  him  the  ends  of  the  threads  of  thought. 

"It  is  due  to  the  peculiarity  of  Mathematics,  which  is  a  chain 
of  inseparable  reasonings,  that  one  part  of  it  can  hardly  be 
studied  to  the  exclusion  of  the  others;  that  in  order  to  under- 
stand the  whole,  only  hard  and  persistent  work,  the  greatest 
perseverance  and  the  greatest  caution,  in  which  all  our  mental 
powers  and  capabilities  have  to  be  brought  into  play,  can  lead 
us  to  the  great  victory  of  the  mind  and  enable  us  to  comprehend 
and  see  the  beauties  of  pure  truths  which  this  magnificent  branch 
of  Science  represents.  To  all  these  peculiarities  is  due  the  fact 
that  only  a  limited  number  of  people  are  capable  of  appreciating 
the  beauties  of  this  oldest  of  all  sciences."  No  fault  has  ever 
been  found  with  Mathematics  by  the  true  student.  He  who 
has  the  courage  to  study  diligently  in  any  line  of  work,  can 
obtain  the  same  results  when  studying  Mathematics  with  the 
same  diligence  and  care.  As  the  drill  will  not  penetrate  the 
granite  unless  kept  to  the  work  hour  after  hour,  so  the  mind 


374  FINKEL'S   SOLUTION  BOOK. 

will  not  penetrate  the  secrets  of  Mathematics  unless  held  long 
and  vigorously  to  the  work.  As  the  sun's  rays  burn  only  when 
concentrated,  so  the  mind  achieves  mastery  in  Mathematics  and 
indeed  in  every  branch  of  knowledge  only  when  its  possessor 
hurls  all  his  forces  upon  it.  Mathematics,  like  all  the  other 
sciences,  opens  its  door  to  those  only  who  knock  long  and  hard. 
No  more  damaging  evidence  can.  be  adduced  to  prove  the  weak- 
ness of  character  than  for  one  to  have  aversion  to  mathematics ; 
for  whether  one  wishes  so  or  not,  it  is  nevertheless  true,  that 
to  have  aversion  for  mathematics  means  to  have  aversion  to 
accurate,  painstaking,  and  persistent  hard  study  and  to  have 
aversion  to  hard  study  is  to  fail  to  secure  a  liberal  education, 
and  thus  fail  to  compete  in  that  fierce  and  vigorous  struggle  for 
the  highest  and  the  truest  and  the  best  in  life  which  only  the 
strong  can  hope  to  secure. 

But  we  do  not  judge  a  painting  by  the  number  of  its  admirers. 
It  is  as  a  rule  the  lowest  kind  of  art  which  attracts  the  largest 
number  of  admirers. 

In  this  practical  world,  in  this  world  of  hard  struggle  for  life, 
where  the  guiding  principle  is  "swim  who  can  and  those  who 
can't  may  drown,"  it  may  not,  perhaps,  be  admissable  to  judge 
of  the  value  of  a  science  by  its  inherent  beauty,  but  rather  by 
the  share  it  contributes  to  the  education  of  our  mental  faculties 
and  by  the  applications  it  finds  in  the  useful  arts  and  sciences 
and  thus  in  what  measure  it  contributes  to  the  civilization  of 
the  world.  He  who  reads  history  with  some  critical  judgment 
cannot  fail  to  notice  that  the  degree  of  civilization  of  a  country 
is  closely  connected  with  the  standard  of  Mathematics  in  that 
country,  and  this  fact  is  attested  by  the  fierce  bidding  for  the 
best  mathematicians  in  the  world  by  such  countries  as  France, 
Russia,  and  Prussia  during  the  latter  part  of  the  last  century. 
Prof.  H.  J.  Stephen  Smith,  of  Oxford,  says,  "I  should  not  wish 
to  use  words  which  may  seem  to  reach  too  far,  but  I  often  find 
the  conviction  ,forced  upon  me  that  the  increase  of  mathematical 
knowledge  is  a  necessary  condition  for  the  advancement  of 
science,  and  if  so,  a  no  less  necessary  condition  for  the  improve- 
ment of  mankind.  I  could  not  augur  well  for  the  enduring  in- 
tellectual strength  of  any  nation  of  men,  whose  education  was 
not  based  on  solid  foundation  of  mathematical  learning  and 
whose  scientific  conception,  or  in  other  words,  whose  notions 
of  the  world  and  of  things  in  it,  were  not  braced  and  girt 
together  with  a  strong  framework  of  mathematical  reasoning." 

Fourier,  one  of  the  greatest  mathematicians  of  France,  on  the 
completion  of  his  great  work  on  Theory  of  Heat,  says,  "Mathe- 
matics develops  step  by  step,  but  its  progress  is  steady  and  cer- 
tain amid  the  continual  fluctuations  and  mistakes  of  the  human 
mind.  Clearness  is  its  attribute,  it  combines  disconnected  facts 
and  discovers  the  secret  bond  that  unites  them.  When  air  and 


GEOMETRY.  375- 

light  and  the  vibratory  phenomena  of  electricity  and  magnetism 
seem  to  elude  us,  when  bodies  are  removed  from  us  into  the 
infinitude  of  space,  when  man  wishes  to  behold  the  drama  of 
the  heavens  that  has  been  enacted  centuries  ago,  when  he  wants 
to  investigate  the  effects  of  gravity  and  heat  in  the  deep,  im- 
penetrable interior  of  our  earth,  then  he  calls  to  his  aid  the  help 
of  mathematical  analysis.  Mathematics  renders  palpable  the 
most  intangible  things,  it  binds  the  most  fleeting  phenomena,  it 
calls  down  the  bodies  from  the  infinitude  of  the  heavens  and 
opens  up  to  us  the  interior  of  the  earth.  It  seems  a  power  of 
the  human  mind  conferred  upon  us  for  the  purpose  of  recom- 
pensing us  for  the  imperfection  of  our  senses  and  the  shortness 
of  our  lives.  Nay,  what  is  still  more  wonderful,  in  the  study 
of  the  most  diverse  phenomena  it  pursues  one  and  the  same 
method,  it  explains  them  all  in  the  same  language,  as  if  it  were 
to  bear  witness  to  the  unity  and  simplicity  of  the  plan  of  the 
universe." 

Mathematics  is  the  very  embodiment  of  truth.  No  true  de- 
votee of  mathematics  can  be  dishonest,  untruthful,  unjust.  Be- 
cause working  ever  with  that  which  is  true,  how  can  one  de- 
velop in  himself  that  which  is  exactly  opposite.  It  would  be 
as  though  one  who  was  always  doing  acts  of  kindness  should 
develop  a  mean  and  groveling  disposition.  Mathematics  there- 
fore has  ethical  value  as  well  as  educational  value.  Its  prac- 
tical value  is  seen  about  us  every  day.  To  do  away  with  every 
one  of  the  many  conveniences  of  this  present  civilization  in 
which  some  mathematical  principle  is  applied,  would  be  to  turn 
the  finger  of  time  back  over  the  dial  of  the  ages  to  the  time 
when  man  dwelt  in  caves  and  crouched  over  the  bodies  of  wild 
beasts. 

The  practical  applications  of  mathematics  has  in  all  ages  re- 
downed  to  the  highest  happiness  of  the  human  race.  It  rears 
magnificent  temples  and  edifices,  it  bridges  our  streams  and 
rivers;  it  sends  the  railroad  car  with  the  speed  of  the  wind 
across  the  continent;  it  builds  beautiful  ships  that  sail  on  every 
sea;  it  has  constructed  telegraph  and  telephone  lines  and  made 
a  messenger  of  something  known  to  mathematics  alone  that  bears 
messages  of  love  and  peace  around  the  globe;  and  by  these 
marvellous  achievements,  it  has  bound  all  the  nations  of  the  earth 
in  one  common  brotherhood  of  man. 

IV.     AXIOMS. 

7.  The  self-evident  first  principles  of  which  mention  was 
made  in  the  previous  section  are  called  axioms. 

Thus,  A  can  not  be  both  B  and  non-./?  at  the  same  time;  A  horse  is  a 
horse;  Two  times  two  are  four;  A  body  in  motion  will  remain  in  motion,, 
unless  ncted  upon  by  some  external  force. 

The  following  are  the  axioms  used  in  mathematics: 


376  FINKEL'S   SOLUTION  BOOK. 


GENERAL  AXIOMS. 

1.  Things  equal  to  the  same  thing  are  equal  to  each  other. 

Thus,  if  A-B  and  B-C,  then  A=C. 

2.  If  equals  are  added  to  equals,  the  sums  are  equal. 

Thus,  if  A—B  and  C—D,  then  A+C=B-\-D. 

3.  If  equals  be  taken  from  equals  the  remainders  are  equal. 

Thus,  if  A-B  and  C=D,  then  A—C=B—D. 

4.  If  equals  be  added  to  unequals  the  sums  are  unequal  in  the  same  order, 
or  sense. 

Thus,  if  A  is  greater  than  B  and  C— D,  then  A-\-C  is  greater  than 
B+D. 

5.  If  equals  be  taken  from  unequals  the  remainders  are  unequal  in  the 
same  sense. 

Thus,  if  A  is  greater  than  B  and  C—D,  then  A—C  is  greater  than 
B— D. 

6.  If  unequals  be  taken  from  equals  the  remainders  are  unequal  in  the 
opposite  sense. 

Thus,  if  A  is  greater  than  B  and  C  is  equal  to  D,  then  C—  A  is  less 
than  D— B. 

7.  If  equals  be  multiplied  by  equals,  the  products  are  equal. 

Thus,  \tA=B  and  C-D,  then  AC=BD. 

8.  If  unequals  be  multiplied  by  equals,  the  products  are  unequal  in  the 
same  sense. 

Thus  if  A  is  greater  than  B  and  C=D,  then  AC  is  greater  than  BD. 

9.  If  equals  be  divided  by  equals,  the  quotients  are  equal. 

Thus,  if  A=B  and  C=D,  then  -£-—-p- 

10.  If  unequals  be  divided  by  equals,  the  quotients  are  unequal  in  the  same 
sense. 

A  B 

Thus,  if  A  is  greater  than  B  and  C=D,  then  —  is  greater  than  -=^ 

11.  If  unequals  be  added  to  unequals,  the  greater  to  the  greater  and  the 
lesser  to  the  lesser,  the  sums  will  be  unequal  in  the  same  sense. 

Thus,  if  A  is  greater  than  B  and  C  greater  than  D,  then  A-\-C  is 
greater  than  B-\-D.  If  m  is  less  than  n  and  p  less  than  q,  then 
m-\-p  is  less  than  n-\-q. 

12.  The  whole  is  greater  than  any  of  its  parts. 

Thus,  if  #!,  #s>  az,a±  are  parts  of  A,  then  A  is  grerter  than  any  of 
the  a's. 

13.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

Thus,  if  0lf  a2,  aa,  «4,  05  are  the  parts  of  ^4,  then  A=a1-\-a9-}-aa-\- 

14.  Magnitudes  which  coincide  with  one  another  are  equal  to  one  another. 

Thus,  if  A  coincides  with  B,  then  A  and  B  are  equal. 

15.  If  of  two  unequal  quantities,  the  lesser  increases  continuously  and  in- 
definitely while  the  other  decreases  continuously  and  indefinitely  they 
must  become  equal  once  and  but  once. 

Thus,  if,  in  the  figure,  the 

line  EF  moves  parallel  0 .  n 

to  itself,  keeping  its  ex-  ~^>^r ** 

tremities    in    AB   and 
MN  and  the  line  GH 
moves  parallel  to  itself    /j^ 
keeping  its  extremities 
in  MN  and  CD,  then 


, 

the  two  lines  are  equal  ^'  *» 

once    and    only    once, 
viz.,   when   both  are  equal   to  the  line  IK. 


GEOMETRY.  37T 

16.  If  of  three  quantities  the  first  is  greater  than  the  second  and  the  second 
greater  than  the  third,  then  the  first  is  greater  than  the  third. 

Thus,  if  A  is  greater  than  B  and  B  greater  than  C,  then  A  is  greater 
than  C. 

17.  Two  straight  lines  can  not  inclose  a  \_finite~\  space. 

V.    ASSUMPTIONS. 

8.  In  addition  to  the  definitions  of  geometrical  magnitudes* 
and  the  above  axioms  the  following  Assumptions,  or  Postu- 
lates, are  needed : 

(a.)     ASSUMPTIONS  OF  THE  STRAIGHT  LINE. 

(i.)  One  and  only  one  straight  line  may  be  passed  through 
every  two  points  in  space;  or,  briefly,  two. points  determine  a 
straight  line. 

(2.)     Two  straight  lines  lying  in  a  plane,  determine  a  point. 

If  the  two  lines  are  parallel,  we  still  say,  for  the  sake  of  generality  and 
in  harmony  with  conventions  adopted  in  modern  geometry,  that  the  two 
lines  intersect  in  a  point,  the  point  infinity.  By  taking  this  view  of  two 
parallel  lines,  many  theorems  are  stated  and  proved  without  exceptions  to 
either  statement  or  proof. 

(3.)  Through  any  point  in  space  a  line  may  be  drawn  and 
revolved  about  this  point  as  a  center  so  as  to  include  any  assigned 
point. 

(4.)  A  straight  line-segment,  or  a  sect,  may  be  produced  so 
as  to  have  any  desired  length. 

(5.)  A  straight  line  is  divided  into  two  parts  by  any  one 
of  its  points. 

(b.)    ASSUMPTIONS  OF  THE  PLANE. 

(i.)     Three  points  not  in  the  same  line  determine  a  plane. 

(2.)  A  straight  line  through  two  points  in  a  plane  lies  wholly 
in  the  plane. 

(3.)  A  plcMie  may  be  passed  through  a  straight  line  and  re- 
volved about  it  so  as  to  include  any  assigned  point  in  spac&. 

(4.)  A  portion  of  a  plane  may  be  produced  to  any  desired 
extent. 

(5.)  A  plane  is  divided  into  two  parts  by  any  of  its  straight 
lines. 

(6.)     A  plane  divides  space  into  two  parts. 

(c.)     ASSUMPTION  OF  PARALLEL  LINES. 

( i.)  Through  a  point  without  a  straight  line,  only  one  straight 
line  can  be  drawn  parallel  to  that  lin&. 

This  assumption  is  a  substitute  for  Euclid's  famous  eleventh 
(also  called  the  twelfth)  axiom  which  reads,  //  a  straight  line 
meet  tzvo  straight  lines  so  as  to  make  the  iivo  interior  angles 

*For  definitions  of  geometrical  magnitudes,  see  Mensuration. 


#78  FINKEL'S   SOLUTION  BOOK. 

on  the  same  side  of  it  taken  together  less  than  two  right  angles, 
these  straight  lines  being  continually  produced  shall  at  length 
meet  on  that  side  on  which  are  the  angles  zvhich  are  less  than, 
two  right  angles. 

An  axiom  must  possess  the  following  properties  :  ( I )  must 
be  self-evident,  (2)  must  be  incapable  of  being  proved  from 
other  axioms.  That  the  above  so-called  axiom  does  not  pos- 
sess the  first  of  these  requisites  is  proved  by  the  fact  that  there 
is  a  dispute  among  mathematicians  as  to  whether  it  is  an  axiom 
or  not.  However,  it  does  satisfy  the  second  criterion  as,  so 
far,  no  valid  proof  of  it  from  other  axioms  has  ever  been  given. 
Many  proofs  have,  indeed  been  given,  but  it  requires  very  little 
thought  to  see  that  these  proofs  are  all  fallacies  of  Petitio 
Principii. 

The  many  attempts  to  give  a  rigorous  and  valid  proof  of  this 
assumption,  for  such  it  is,  has  redounded  to  the  eternal  glory  of 
geometry  in  that  not  only  is  Euclidean  Geometry  preserved  in 
all  its  original  purity  and  integrity  but  other  geometries  equally 
cogent  and  consistent  have  been  created. 

The  subject  is  too  abstruse  for  my  present  purpose  and  so  I 
shall  do  nothing  more  than  show  the  point  of  departure  of  these 
geometries. 

1.  Let  AB  be  a 
given  straight  line, 
and   P    the    given 
point. 

2.  Through     P 
draw  any   number 
of  lines. 

3.  These    lines, 
in   relation   to  the 

given  line,  divide  themselves  into  two  classes,  viz.,  CUTTING  and 

NON-CUTTING. 

Now  of  the  class,  non-cutting,  how  many  lines  are  there?  On 
the  answer  to  this  question  "hangs  all  the  law  and  the  prophets." 
A  priori,  three  answers  are  possible,  viz.,  none,  one,  many.  If 
we  say  "none,"  we  have  Spherical  Geometry;  if  we  say 
"one,"  we  have  Euclidean  Geometry;  if  we  say  "more  than 
one,"  we  have  Pseudo- Spherical  Geometry. 

It  is  true  that  the  answer,  "one,"  is  the  answer  that  is  usually 
insisted  upon  as  being  the  only  possible  answer.  But  this  an- 
swer is  based  upon  experience  and  is  not,  therefore,  a  priori. 

In  these  geometries,  the  properties  of  figures  are  studied,  which 
figures  lie  in  space,  or  surfaces,  possessing  the  property  that 
the  product  of  the  principal  radii  of  curvature  at  every  point 
of  the  surfaces  shall  be  constant.  If  this  product  is  positive,  the 
surface  is  spherical  and  the  geometry  treating  of  the  figures  of 
this  surface  is  Spherical  Geometry;  if  this  product  is  0,  the 


GEOMETRY.  379 

•surface  is  a  plane,  and  the  geometry  treating  of  the  properties 
of  figures  lying  in  this  surface  is  the  ordinary  Euclidean  Geom- 
etry ;   if  this  product  is  negative,  the  surface  is  pseudo-spherical 
and   the   geometry   treating  of   the   properties   of   this   space   is_ 
Pseudo-Spherical  Geometry. 

In  the  above  discussion,  it  has  been  assumed  tacitly  that  the 
measure  of  a  distance  remains  everywhere  the  same.  Professor 
Felix  Klein  has  shown  that  if  this  be  not  the  case  and  if  the 
lav/,  of  measurement  of  distance  be  properly  chosen,  we  can 
obtain  three  systems  of  plane  geometry  analogous  to  the  three 
systems  mentioned  above.  These  are  called  respectively  Ellip- 
tic, Parabolic,  and  Hyperbolic  Geometries.  They  mean 
lacking,  equaling,  and  exceeding.  Instead  of  the  above  terms 
given  by  Klein,  we  often  meet  Riemannian,  Euclidean, 
and  Isobatschevskian  (or  Gaussian)  from  Riemann,  Euclid, 
and  Lobatschevsky  and  Gauss,  —  mathematicians  who  first  set 
forth  clearly  the  properties  of  the  space-forms.  These  geom- 
etries refer  to  hyper-space  of  two  dimensions  and  are  called 
collectively  non- -Euclidean  Geometry. 

The  notion  of  hyper-space  of  two  dimensions  naturally  sug- 
gested the  question  as  to  whether  there  are  different  kinds  of 
hyper-s^>ace  of  three  or  more  dimensions.  Riemann  showed  that 
there  are  three  kinds  of  hyper-space  of  three  dimensions  having 
properties  analogous  to  the  three  kinds  of  hyper-space  of  two 
dimensions  already  discussed.  These  hyper-spaces  are  differ- 
entiated by  the  test  whether  at  every  point  no  geodetical  surface, 
or  one  geodetical  surface,  or  a  fasciculus  of  geodetical  surfaces 
can  be  drawn  parallel  to  a  given  surface,  a  geodetical  surface 
being  defined  as  such  that  every  geodetic  line  joining  any  two 
points  on  it  lies  wholly  on  the  surface.  The  student  who  would 
pursue  the  subject  should  read  Dr.  Halsted's  excellent  transla- 
tions of  Lobatschevsky  and  Bolyai,  the  Lectures  and  Addresses 
of  Clifford  and  Helmholz,  Ball's  article  on  Measurement  in  the 
Encyclopedia  Britannica,  Professor  Schubert's  Essay  on  the 
Fourth  Dimension,  Russell's  Foundations  of  Geometry,  and  after- 
wards the  monographs  of  Riemann,  Klein,  Newcomb,  Beltrami, 
and  Killing.  For  a  full  bibliography  of  the  literature  of  the 
subject  up  to  the  time  of  its  publication,  see  Bibliography  of  Non- 
Euclidean  Geometry,  by  Dr.  Halsted,  American  Journal  of 
Mathematics. 

(d.)     ASSUMPTION  OF  THE  CIRCLE. 

(i.)  A  circle  may  be  constructed  with  any  point  as  center, 
and  with  a  radius  equal  to  any  given  sect. 

(2.)     A  circle  has  but  one  center. 

(3).  All  radii  of  the  same  circle  arc  equal,  and,  hence  all 
diameters  of  the  same  circle  are  equal. 


380  FINKEL'S   SOLUTION   BOOK. 

(4.)  //  an  unlimited  straight  line  passes  through  a  point 
within  a  circle,  it  must  cut  the  circumference  at  least  twice. 

That  it  can  not  cut  the  circumference  more  than  twice  is  a 
theorem. 

The  region  within  a  circle  is  defined  as  that  from  any  point 
of  which  no  tangents  can  be  drawn'to  the  circle. 

(5.)  //  one  circumference  intersects  another  once,  it  inter- 
sects it  again. 

(e.)    ASSUMPTION  OF  THE  SPHERE. 

(i.)  A  sphere  may  be  constructed  with  any  point  as  center, 
and  with  a  radius  equal  to  any  given  sect. 

(2.)     A  sphere  has  but  one  center. 

(3.)  All  radii  of  the  same  sphere  are  equal,  and,  hence  all 
diameters  of  the  same  sphere  are  equal. 

(4.)  //  an  unlimited  straight  line  passes  through  a  point 
within  a  sphere,  it  must  cut  the  surface  at  least  twice. 

(5.)  If  an  unlimited  plane  or  if  a  spherical  surface,  intersects 
a  spherical  surface,  it  must  intersect  it  in  a  closed  line. 

(/)     ASSUMPTION  OF  MOTION. 

(i.)  A  figure  may  be  moved  from  one  position  in  three  di- 
mensional space  to  any  other  position  in  the  same  space  without 
altering  the  size  or  shape  of  the  figure. 

By  this  we  mean  that  a  figure  may  be  picked  up,  turned  over  in  any 
way,  and  moved  to  any  other  position  in  space  without  changing  the  size 
or  shape  of  the  figure.  The  proof  of  many  theorems  in  geometry  depends 
upon  this  assumption. 

(2.) A  figure  may  be  moved  about  in  space  while  one  of  its 
points  remains  fixed. 

Such  movement  is  called  "  rotation  about  a  center,"  the  center  being 
the  fixed  point 

(3.)  A  figure  may  be  moved  about  in  space  while  two  of  its 
points  remain  fixed. 

Such  movement  is  called  "  rotation  about  an  axis,"  the  axis  being  the 
line  determined  by  two  fixed  points. 

In  the  higher  mathematics  and  in  Physics  and  other  natural 
sciences  other  assumptions  are  needed. 

VI.     ON  LOGIC. 

9.  On  I/ogic. —  As  a  preliminary  to  the  study  of  geometry 
a  short  discussion  of  the  Methods  of  Reasoning  will  be  of  value. 

In  geometry  we  are  concerned  with  propositions  about 
space  relations.  Ideas  are  images  of  an  object  formed  by  the 
mind.  Words  are  the  spoken  or  written  signs  of  ideas. 

10.  A  judgment  is  an  act  of  the  mind  affirming  a  relation 
between  two  objects  of  thought  by  means  of  their  conceptions. 


U2/V~t^~ 


GEOMETRY.  381 

11.  A  proposition  is  a  judgment  expressed  in  words. 
For  example,  take  the  ideas  represented  by  "all  mushroons" 

and  "things  good  to  eat,"  posit  these  ideas  in  the  mind  and  dis- 
cern the  agreement  or  disagreement  of  these  two  ideas,  then 
express  the  agreement  or  disagreement  in  words.  It  comes  out 
thus, 

"All  mushroons  are  things  good  to  eat." 

Our  senses  are  the  instruments  by  which  the  qualities  of  a 
mushroon  are  made  known  to  us.  Having  found  this  mushroon 
good  to  eat,  and  this  one,  and  this  one,  and  so  on,  together  with 
the  experience  of  the  race,  we  arrive  at  the  conclusion,  by  in- 
ductive inference,  that  "all  mushroons  are  good  to  eat."  It  must 
be  borne  in  mind  that  by  induction  we  gain  no  certain  knowledge. 
If  the  observation  of  a  number  of  cases  shows  that  alloys  of 
metals  fuse  at  lower  temperatures  than  their  constituent  metalsr 
we  may  with  more  or  less  probability  draw  the  general  inference 
that 

All  alloys  melt  at  a  lower  temperature  than  their  constituent 
metals. 

But  this  can  never  rise  to  the  rank  of  an  absolutely  certain 
law  until  all  possible  cases  have  been  examined.  Not  one  of 
the  inductive -truths  which  men  have  established,  or  think  they 
have  established,  is  really  safe  from  exception  or  reversal. 
Lavoisier,  when  laying  the  foundations  of  chemistry,  met  with 
so  many  instances  tending  to  show  the  existence  of  oxygen  in 
all  acids  that  he  adopted  the  general  conclusion  that  all  acids 
contain  oxygen,  yet  subsequent  experience  has  shown  this  to 
be  false.  Like  remarks  may  be  made  concerning  all  other  in- 
ductive inferences,  the  method  never  leading  to  absolute  certainty. 

12.  The  Powers  of  the  Mind  engaged  in  knowledge  are  the 
following  three,  viz., 

1 i )  The  Power  of  Discrimination, 

(2)  The  Power  of  Detecting  Identity,  and 

(3)  The  Power  of  Retention. 

13.  The  Laws  of  Thought  are  the  following  three,  viz., 

1 i )  The  Law  of  Identity ;  as,  That  which  is,  is. 

(2)  The  Law  of  Contradiction ;  as,  A  thing  cannot  both 
be  and  not  be  at  the  same  time. 

(3)  The  Law  of  Duality;  as,  A  thing  must  either  be  or 
not  be. 

To  these  some  logicians  add  a  fourth  called  the  "Law  of  Suf- 
ficient reason ;"  Every  effect  has  a  cause. 

14.  When  we  join  terms  together  we  make  propositions; 
when  we  join  propositions  together  we  make  an  argument,  or 
piece  of  reasoning. 


382  FINKEL'S  SOLUTION  BOOK. 

15.  Terms.     A  concrete  term  has  two  meanings,  viz.,  (i) 
things  to  which  the  term  applies,  and    (2)    the  qualities  of  those 
things  in  consequence  of  which  the  term  is  applied.     The  num- 
ber of  different  things  to  which  a  term  is  applied  is  called  its 
extension,  while  the  number  of  qualities  implied  is  called  its 
intension. 

For  example,  "table"  has  a  larger  "extension"  than  "round 
table"  for  the  former  term  applies  to  a  larger  number  of  objects ; 
the  latter  has  the  greater  "intension"  for  it  includes  all  the  quali- 
ties that  the  term  "table"  does  and  the  additional  quality  "round." 

The  word  "term"  comes  from  the  Latin  terminus,  meaning 
end  and  is  so  called  because  it  forms  one  end  of  a  proposition. 

1 6.  Propositions.     Every  •  proposition   is   composed  of   a 
subject,  (  Lat.,  sub,  under,  and  jectum,  laid),  a  copula,,  and  a 
predicate  (Lat.  praedicare,  to  assert). 

In  the  proposition,  "All  mushroons  are  things  good  to  eat," 
"all  mushroons"  is  the  subject,  "are"  is  the  copula,  and  "things 
good  to  eat"  is  the  predicate. 

Of  the  kinds  of  propositions  we  have 

(i)  Categorical;  As  A  is  B.  A  is  not  B;  (2)  Con- 
ditional; as,  If  a  triangle  is  equiangular,  it  is  equilateral. 

Conditional  Propositions  are  divided  into  two  classes, 
viz.,  Hypothetical  and  Disjunctive.  The  following  is  a 
disjunctive  proposition: 

A  is  either  B  or  C. 
Of  the  Categorical  Propositions  we  have, 

A.     The  Universal  Affirmative;   as,    All    horses    are 

animals. 
E.     The  Particular  Affirmative;  as,  Some  animals  are 

horses. 

I.     The  Universal  Negative;  as,  No  horses  are  cows. 
O.     The  Particular  Negative;  as,  Some  animals  are  not 

horses. 

Every  proposition  which  expresses  accurately  a  thought,  can 
be  reduced  to  one  of  the  above  forms,  though  the  reduction  in 
many  cases  is  not  apparent.  For  example, 

Parallel  lines  never  meet,  reduces  to 
Parallel  lines  are  lines  which  never  meet. 

The  hypothetical  proposition,  "If  gunpowder  be  damp,  it  will 
not  explode"  reduces  to,  "Damp  gunpowder  will  not  explode." 

When  we  make  a  statement  about  all  the  objects  which  can 
be  included  under  a  term,  we  use  the  term  UNIVERSALLY,  as 
logicians  say,  that  is  to  say,  THE  TERM  is  DISTRIBUTED.  In  the 
proposition,  "all  men  are  mortal,"  the  term  "men"  is  distributed, 
because  the  little  word  "all"  indicates  that  the  statement  applies 


GEOMETRY.  383 

to  any  and  every  man.     But  THE  PREDICATE  "mortal"  is  ONLY 

TAKEN  PARTICULARLY  AND  IS  NOT  DISTRIBUTED. 

Therefore,  we  see  that  a  UNIVERSAL  AFFIRMATIVE  DISTRIB- 
UTES ITS  SUBJECT  BUT  NOT  ITS  PREDICATE. 

As  a  universal  negative  proposition  take,  "No  sea-weed  is  a 
flowering  plant."  The  subject  "sea-weed"  is  distributed.  If 
there  could  be  found  a  single  flowering  plant  which  is  a  sea- 
weed, then  the  proposition  would  not  be  true.  Hence  the  predi- 
cate is  also  distributed. 

Hence,  THE  UNIVERSAL  NEGATIVE  PROPOSITION  DISTRIBUTES  ITS 

SUBJECT  AND  ITS   PREDICATE. 

No  difficulty  is  experienced  in  seeing  that  the  particular  affirma- 
tive distributes  neither  its  subject  nor  its  predicate,  and  that  the 

PARTICULAR    NEGATIVE    DISTRIBUTES    ITS    PREDICATE    BUT    NOT    ITS 
SUBJECT. 

In  the  absence  of  any  knowledge  to  the  contrary,  the  word 
"some,"  in  the  particular  affirmative  and  particular  negative,  must 
be  taken  to  mean  "SOME  AND  IT  MAY  BE  ALL/' 

17.  Tlie  I/a w  of  Converse.     Two   propositions   are   the 
converse  of  each  other  when  the  subject  of  one  is  the  predicate 
of  the  other.     Thus, 

"Equilateral  triangles  are  equiangular,  "(direct). 
Equiangular  triangles  are  equilateral,  (converse). 

It  does  not  follow  that  because  a  proposition  is  true  its  con- 
verse will  also  be  true.  Thus,  "All  regular  polygons  are  equi- 
lateral (direct)  ;  all  equilateral  (polygons)  are  regular,  (con- 
verse). This  last  is  not  true.  The  converse  of  all  definitions 
are  true. 

Whenever  three  theorems  have  the  following  relations,  their 
converses  are  true : 

1.  If  it  is  known  that  when  A  >  B,  then  x  *>  y,  and 

2.  If  it  is  known  that  when  A  =  B,  then  x  —  y,  and 

3.  If  it  is  known  that  when  A  <  /?,  then  x  < y, 
then  the  converse  of  each  of  these  is  true. 

For 

lx.  If  x  >y,  then  A  cannot  equal  B  and  A  cannot  be  less 
than  B  without  violating  2  or  3;  .'.  A  >  B.  (Converse  of  1.) 

218  If  x  —  y,  then  A  cannot  be  greater  than  B  and  A  cannot 
be  less  than  B  without  violating  1  or  3;  .*.  A=B.  (Converse 
of  2.) 

3t.  If  x  < y,  then  A  cannot  be  greater  than  B  and  A  cannot 
be  equal  to  B  without  violating  1  or  2;  .'.  A  <  B.  (Converse 
of  3.) 

1 8.  The    opposite  of  a  proposition  is  formed  by  stating  the 
negative  of  its  hypothesis  and  conclusions.     Thus, 

If  A  —  B,  then  C  =  D  (Direct.) 

If  A  is  not  equal  B,  then    C  is  not  equal  D.     (Opposite.) 


384  FINKEL'S  SOLUTION  BOOK. 

19.  //  the  direct  proposition  and  its  converse  are  true,  the 
opposite  proposition  is  true;   and  if  a  direct  proposition  and  its 
opposite  are  true,  the  converse  proposition  is  true.    Thus, 

1.  It  A=B,  C  =  D.     (Direct.) 

If  C  =  D,  A  =  B.     (Converse.) 

If  A  is  not  equal  to  B,-C  is  not  equal  to  D      (Opposite.) 

2.  If  A  =  B,  C  =  D.     (Direct.) 

If  A  is  not  equal  to  B,  C  is  not  equal  to  D.     (Opposite.) 
Then,  if  C  =  D,  A  =  B.     (Converse.) 

20.  Methods  of  Reasoning.   There  are  two  methods  of 
reasoning,  viz.,  the  Inductive  and  the  Deductive. 

The  Inductive  Methodis  used  in  reaching  a  general  truth 
or  principle  by  an  examination  and  comparison  of  particular 
facts.  Thus,  This  apple  is  equal  to  the  sum  of  all  its  parts,  this 
piece  of  crayon  is  equal  to  the  sum  of  all  its  parts,  this  orange 
is  equal  to  the  sum  of  all  its  parts,  and  so  with  peaches,  pears,, 
balls,  pebbles,  slates,  knives,  and  chairs. 

Therefore,  the  whole  of  any  object  is  equal  to  the  sum  of  all 
its  parts,  or  the  whole  is  equal  to  the  sum  of  all  its  parts.  This 
is  inductive  reasoning. 

The  Deductive  Method  is  used  in  reaching  a  particular 
truth  or  principle  from  general  truths  or  principles.  Thus. 

All  animals  suffer  pain. 
Flies  are  animals. 
Therefore,  flies  suffer  pain. 

21.  Tie  Syllogism.   When  we  compare  propositions  we 
reason.     Deriving  a  third  proposition  from  two  given  proposi- 
tions is  called  syllogistic  reasoning,  or  Deductive  Rea- 
soning.   Thus, 

1.  All  English  silver  coins  are  coined  at  Tower 

Hill. 

2.  All  sixpences  are  coined  at  Tower  Hill. 
Therefore,  All  sixpences  are  English  silver  coins. 

The  last  proposition  is  called  the  conclusion,  the  other  two 
propositions  are  called  premises,  and  the  three  together  the 
syllogism. 

Again, 

All  electors  pay  rates.  A. 

No  paupers  pay  rates.  E. 

Therefore,  no  paupers  are  electors.       E. 

From  the  examples  given,  we  see  that  there  are  only  three 
terms  or  classes  of  things  reasoned  about ;  in  the  first  example 
the  three  terms  are  "All  English  silver  coins/'  "Tower  Hill,"  and 
"  all  sixpences."  Of  these,  the  class,  "  English  silver  coins," 
does  not  occur  in  the  conclusion.  It  is  used  to  enable  us  to 
compare  together  the  other  two  classes  of  things.  It  is  called 


RENE    DESCARTES. 


GEOMETRY.  385 

the  middle  term.  (Things)  "coined  at  Tower  Hill,"  is  called 
the  major  term  for  the  reason  that  it  has  the  larger  exten- 
sion, and  "sixpences,"  the  subject  of  the  conclusion,  is  called  the 
minor  term  of  the  syllogism,  for  the  reason  that  it  has  a  lesser 
extension  than  the  subject  of  the  conclusion. 

The  premise  in  which  the  "major  term"  is  found  is  called  the 
major  premise,  and  the  one  in  which  the  minor  term  is  found 
is  called  the  minor  premise. 

Hence,  the  middle  term  is  always  the  term  not  found 
in  the  conclusion ;  the  major  term  is  the  predicate  of 
the  conclusion ;  and  the  minor  term  is  the  subject  of 
the  conclusion. 

Suppose  that  the  two  premises  and  the  conclusion  of  the  last 
syllogism  be  varied  in  every  possible  way  from  affirmative  to 
negative,  from  universal  to  particular  and  vice  versa. 
.  Each  proposition  can  be  converted  into  four  different  propo- 
sitions and  each  one  of  these  four  may  be  compounded  with  any 
one  of  the  other  two.  Hence  the  number  of  changes  (called 
moods)  is  4  X  4  X  4  —  64.  These  moods  may  be  still  further 
varied,  if  instead  of  the  middle  term  being  the  subject  of  the 
first  and  the  predicate  of  the  second,  this  order  may  be  reversed, 
or  if  the  middle  term  the  subject  of  both,  or  the  predicate  of 
both.  In  this  way  we  see  that  for  each  of  the  sixty-four  moods 
we  get  four  syllogisms  called  figures. 

Of  the  sixty-four  moods,  there  are  altogether  nineteen 
moods  of  the  syllogism  that  are  admissible. 

22.  Rules  of  the  Syllogism.  To  find  out  whether  an 
argument  is  valid  or  not,  we  must  examine  it  carefully  to  ascer- 
tain whether  it  agrees  with  certain  rules  discovered  by  Aristotle. 
Modern  logicians  have  to  some  extent  broken  away  from  these 
rules.  Without  going  into  the  matter  in  detail  we  state  these 
rules. 

I.  Every  syllogism  has  three  terms  and  only  three. 
These  terms  are  called  the  the  major  term,  the  minor  term,  and 
the  middle  term. 

II.  Every  syllogism  contains  three  and  only  three 
propositions. 

III.  The  middle  term  must  be  distributed  once  at 
least  in  the  premises  and  must  not  be  ambiguous. 

Some  animals  are  flesh-eating. 
Some  animals  have  two  stomachs. 
No  conclusion  can  be  drawn. 


But  if  we  say, 

Some  animals  are  flesh-eating, 
All  animals   consume   oxygen,    we 
can  say 

Therefore,  some  animals  consuming  oxygen  are  flesh-eating. 


FINKEL'S  SOLUTION  BOOK. 


IV.  If  both  premises  are  negative  no  conclusion 
can  be  drawn. 

For,  from  the  statements  that  two  things  disagree  with  a  third, 
no  proof  of  agreement  or  disagreement  can 
be  established.     Thus  the  following  is  incon- 
clusive, 

No  Americans  are  slaves. 

No  Turks  are  Americans. 

V.  If  both  premises  are  particular 
no  conclusion  can  be  drawn. 

Thus  the  following  are  inconclusive: 

Some  Americans  are  ignorant. 
Some  Europeans  are  ignorant. 

Some  laws  are  unjust. 
Some  men  are  unjust. 

VI.  No  term  must  be  distributed  in  the  conclusion 
which  was  not  distributed  in  the  premises. 

From 

Some  animals  eat  flesh. 

All  animals  consume  oxygen. 

We  must  conclude  that  some  things  that 
consume  oxygen  eat  flesh. 


VII.  If  one  premise  be  negative   the   conclusion 
must  be  negative. 

Thus  from 

All  negroes  are  dark. 
No  American  is  dark. 

We  draw  the  conclusion 

No  American  is  a  negro. 

VIII.  If  either  premise  is  particu- 
lar the  conclusion  must  be  particular. 

Thus, 

All  negroes  are  black. 
Some  horses  are  black. 

Therefore,  some  horses  are  not  negroes. 


23.  logical  Fallacies.  Logical  Fallacies  result  from  our 
neglect  to  observe  the  rules  of  logic.  They  occur  in  the  mere 
form  of  the  statement,  that  is,  in  dictione,  as  it  is  known  in  logic. 


GEOMETRY.  387 

There  are  four  purely  logical  fallacies,  viz., 

1.  Fallacy  of  four  terms   (Quaternio  Terminorum),  — 

Violation  of  Rule  I. 

2.  Fallacy  of  undistributed  middle,  —  Violation  of  Rule 

III. 

3.  Fallacy  of  illicit  process,  of  the  major  or  minor  term. 

—  Violation  of  Rule  VI. 

4.  Fallacy  of  negative  premises.  —  Violation  of  Rule 

IV. 
There  are  six  semi-logical  fallacies,  viz., 

1.  Fallacy  of  Equivocation. 

2.  Fallacy  of  Amphibology. 

3.  Fallacy  of  Composition. 

4.  Fallacy  of  Division. 

5.  Fallacy  of  Accent. 

6.  Fallacy  of  Figure  of  Speech. 

In  addition  to  these  logical  fallicies  there  are  seven  Material 
Fallacies  (extra  dictionem)tha.t  is,  fallacy  in  the  matter  of 
thought,  viz., 

1.  Fallacy  of  Accident. 

2.  The  Converse  Fallacy  of  Accident. 

3.  The  Irrelevant  Conclusion. 

4.  The  Petitio  Principii. 

5.  The  Fallacy  of  the  Consequent  or  Non-sequitur. 

6.  The  False  Cause. 

7.  The  Fallacy  of  Many  Questions. 
We  will  illustrate  some  of  these  fallacies. 

Light  is  contrary  to  darkness. 
Feathers  are  light. 
.-.  Feathers  are  contrary  to  darkness. 

The  middle  term,  "light,"  has  two  different  meanings  in  the 
premises.  We  have,  therefore,  four  terms  instead  of  three,  which 
violates  Rule  I.  When  the  middle  term  is  ambiguous,  the  fal- 
lacy is  known  as  the  ambiguous  middle. 

Every  country  under  a  tyranny  is  distressed. 
This  country  is  distressed. 

.-.  This  country  is  under  a  Tyranny.  —  Fallacy  of  Undis- 
tributed Middle. 

All  moral  beings  are  accountable. 
No  brute  is  a  moral  being. 

.-.  No  brute  is  accountable.  —  Fallacy  of  the  Illicit  Process 
of  the  Major  Term. 

Some  men  are  not  just. 
No  angel  is  a  man. 
.-.  Some  angels  are  not  just. —  Fallacy  of  Negative  Premises. 


388  FINKEL'S   SOLUTION  BOOK. 


EXAMPLES. 

Seven  is  one  number. 
Two  and  five  are  seven. 
.-.  Two  and  five  are  one  number.  —  Fallacy  of  Division. 

Three  and  four  are  two  numbers. 
Seven  is  three  and  four. 
.-.  Seven  is  two  numbers.  —  Fallacy  of  Composition. 

The  duke  yet  lives  that  Henry  shall  depose.  —  Fallacy  of 
Am  philology. 

The  conclusion  depending  upon  the  interpretation  of  the  mean- 
ing of  this  proposition  is  doubtful. 

A  hero  is  a  lion. 
A  lion  is  a  quadruped. 
.-.A  hero  is  a  quadruped.  —  Fallacy  of  Figure  of  Speech. 

Thieves  are  dishonest ; 
But  thieves  are  men; 
.-.  All  men  are  dishonest.  —  Fallacy  of  Accident. 

24.  How  to  Prepare  a  L,esson  in  Geometry.  In  be- 
ginning the  study  of  geometry,  great  care  should  be  taken  to  grasp 
a  correct  notion  of  the  definitions  and  illustrations.  The  defi- 
nitions, axioms,  and  assumptions  are  the  foundation  on  which 
rests  the  magnificent  structure  of  geometry.  The  definitions 
should  be  committed  to  memory,  only  committing  them,  however, 
as  they  occur  in  the  prosecution  of  the  study.  Make  haste  slowly 
at  first ;  one  proposition  per  lesson  for  the  first  three  lessons 
is  quite  sufficient;  and  two  propositions  may  be  taken  at  a  les- 
son for  the  next  seven  or  eight  lessons.  After  this,  if  the  work 
is  thoroughly  in  hand  three  propositions  together  with  several 
originals  should  constitute  a  lesson. 

In  the  preparation  of  the  lesson,  the  student  should  carefully 
read  the  proposition  so  as  to  get  its  full  meaning.  After  the 
meaning  of  the  proposition  is  understood,  carefully  follow  the 
demonstration  in  the  book,  never  leaving  a  statement  made  in 
the  demonstration  until  it  is  thoroughly  understood.  At  first, 
it  may  be  necessary  to  repeat  this  two  or  three  times,  perhaps 
oftener.  After  the  given  demonstration  is  thoroughly  under- 
stood, close  the  book,  draw  a  figure  on  paper  or  a  slate,  and 
write  out  a  demonstration  of  your  own.  Compare  your  demon- 
stration with  the  one  in  the  book,  and  make  such  corrections  as 
are  necessary. 

By  carefully  observing  this  method,  it  will  be  a  comparatively 
short  time  until  one  reading  of  the  lesson  will  generally  suffice 
for  the  necessary  preparation.  The  theorem  should  always  be 


LEONHARD  EULER. 


GEOMETRY. 


389 


committed  to  memory,  the  demonstration  never.  It  is  not  a  bad 
practice  to  commit  the  proposition  exactly  as  it  is  stated  in  the 
book,  for,  as  a  general  thing  the  author  has  put  much  time  on 
the  statement  of  each  proposition  endeavoring  to  reduce  it  to 
its  simplest  and  most  elegant  form,  and  upon  this  work,  the 
student,  as  a  rule,  can  not  improve. 

In  conducting  the  recitations,  no  books  should  be  allowed  to 
be  consulted.  The  propositions  should  be  assigned  by  stating 
them  in  part  or  in  full  to  the  students  called  upon  to  recite.  The 
students  so  called  upon,  should  go  to  the  board  and  draw  as 
neat  and  accurate  figure  as  possible,  accurate  figures  often  sug- 
gesting truths  not  revealed  by  carelessly  constructed  figures.  It 
is  generally  best  not  to  require  any  part  of  the  demonstration  to 
be  written  out,  unless,  indeed,  it  includes  long  and  complicated 
algebraic  equations.  In  reciting,  if  it  is  convenient,  the  student 
should  step  to  the  board  and,  using  a  pointer  in  referring  to  the 
various  parts  of  his  figure,  observe  the  following  order  in  the 
discussion  of  the  theorem : 

I.     Statement  of  the  Theorem.    Here  give  an  accu- 
rate statement  of  the  theorem  to  be  demonstrated. 
II.     Given.     Here  state,  with  reference  to  the  figure  con- 
structed whatever  is  given  by  the  theorem. 

III.  To  Prove.     Here  state  the  exact  conclusion  to  be  de- 

rived from  what  is  given. 

IV.  Proof.     Here  set  forth,  in  logical  order  the  statements 

to  prove  the  conclusion  just  asserted. 

The  validity,  limitations,  and  general  application  of  the  the- 
orem may  then  be  discussed  by  the  class. 

Corollaries  coming  under  the  various  theorems  in  the  lesson 
may  be  assigned  to  students  other  than  those  demonstrating  the 
theorems.  The  proof  of  a  corollary  is  usually  simple,  but  its 
proof  should  be  given  with  the  same  care  and  accuracy. 

We  will  now  illustrate  what  we  have  said  by  a  few  proposi- 
tions. The  student  should  have  one  of  the  following  excellent 
texts : 

Halsted's  Elements  of  Geometry.. 

Beman  and  Smith's  Plane  and  Solid  Geometry. 

Phillips  and  Fisher's  Elements  of  Geometry. 

Wentworth's  Plane  and  Solid  Geometry. 


390 


FINKEL'S  SOLUTION  BOOK. 


PLANE    GEOMETRY. 


BOOK  I. 


ANGLES  AND   STRAIGHT   LINES. 


I.     Theorem. 


PROPOSITION  I. 

All  straight  angles  are  equal. 


C 

I 


.B 


I. 


II. 
III. 


E 

I. 


.F 


II.     Given        any  two  straight  angles  ACB  and  DEF. 
III.     To  prove  /_  ACB  =  /  DEF. 

f  1.     Apply  /  ACB  to  the  /  DEF,  so  that  the 

vertex  C  shall  fall  on  the  vertex  E. 
(First  assumption  of  motion.) 

2.  Then  revolve  CB  so  that  it  contains  the 

point  F. 
(Third  assumption  of  the  straight  line.) 

3.  Then  CA  will  coincide  with  ED. 

(First  assumption  of  a  straight  line  and 

Law  of  Identity?) 
14.     .'.  tACB  =  /_DEF.     Axiom  10. 

Corollary  1.     All  right  angles  are  equal. 
A  A 


IV.    Proof. 


C 
Given 


B 


C 


any  two  right  angles  ACB  and  A  C' Bf . 
To  prove  /_  ACB  =  /  ACB'. 


1.     All  straight  angles  are  equal.     Prop.  I. 

IV      Proof.  \^'     ^-ACB.  and  /  ^C'ff  are  each  the  half  of  a 

straight  angle.     By  definition. 
3.     .-.  l_ACB  =  tAC'B'.     Axiom?. 


GEOMETRY. 


391 


I.     Cor.  2.     The  angular  units,  degree,  minute,  and  sec- 
ond have  constant  values. 

II.     Given         a  degree  angle. 
III.     To  prove  that  it  is  a  constant  magnitude. 

1.     A    constant    magnitude    is    a  magnitude 

By 


IV.    Proof. 


whose  value  is  always  the  same, 
def. 

A  straight  angle  is  a  magnitude  whose 
value  is  always  the  same.     By  Prop.  I. 

1.  .'.A  straight  angle  is  a  constant  mag- 

nitude. 

2.  A  degree  angle  is  one  hundred  eight- 

ieth part  of  a  straight  angle.  By 
def. 

3.  .'.A  degree  angle  is  a  constant  mag- 

nitude. By  Aristotle's  Dictum, — 
Whatever  may  be  predicated  of  a 
whole  may  be  predicated  of  a  part. 

In  like  manner,  we  can  prove  that  minute-angles  and  second- 
angles  are  constants. 

I.     Cor.  3.     Complements  of  equal  angles  are  equal. 


IV.    Proof. 


II.     Given        the  two  equal  angles  CBD  and  CB'tf  and 
their  complements  ABC  and  A B' C ',  respectively. 

III.     To  prove  that  /  ABC  =  /  A'B'C. 

/  ABC  =  the  difference  between  a  rt.  / 

and  /  CBD.     By  def.  of  comp. 
/_  A'B'C'  =  the  difference  between  a  rt.  / 
"  and  /  C'B'jy.     By  def.  of  comp. 
But  /  CBD  =  /  C'B'Df.     By  hypothesis. 
/.  /  ABC  =  /  A'B'C.     By  Axiom  1. 

I.     Cor.  4.     Supplements  of  equal  angles  are  equal. 
(Proof  same  as  above.) 

I.  Cor.  5.  At  a  given  point  in  a  given  line,  one  perpen- 
dicular, and  only  one,  can  be  erected  in  the  same 
plane. 


392 


FINKEL'S  SOLUTION  BOOK. 


II.     Given 


,-r  '/> 

CD  perpendicular  to  AB  at  P. 


IV.    Proofs 


III.     To  prove  that  no  other  perpendicular  can  be  drawn  to 
AB  at  P  in  the  same  plane. 

fl.     Suppose   that  another  perpendicular  EF 
could  be  drawn. 

2.  Then  /  BPE  would  be  a  rt.  /.    By  def.  of 

perpendicular. 

(If  two  lines  meet  and  form  a  rt.  L,  each  is  said  to  be  per- 
pendicular to  the  other.) 

3.  But  /  BPC  is  a  rt.  angle. 

(Since  CD  is  perpendicular  to  AB.) 

4.  /;  2  BPE  would  equal  /  BPC.    Prop.  I., 

Cor.  1. 

(All  right  angles  are  equal.) 

5.  But  this  is  impossible.     By  Axiom  8. 

(The  whole  is  greater  than  any  of  its  parts.) 

6.  .'.  The   supposition  of  step  1  is  absurd, 

and  a  second  perpendicular  is  impossi- 
ble.    Q.  E.  D. 

Remark.  In  this  demonstration,  we  have  used  what  is  called 
the  Indirect  Method,  or  reductio  ad  absurdum  which  means 
a  reduction  to  an  absurdity,  as  distinguished  from  the  Direct 
Method  used  in  the  other  proofs.  Jevons  in  his  Principles  of 
Science,  Vol.  I,  p.  96,  says,  "Some  philosophers,  especially  those 
of  France,  have  held  that  the  Indirect  Method  of  Proof  has  a 
certain  inferiority  to  a  direct  method,  which  should  prevent  our 
using  it."  He  goes  on  to  show  that  the  method  is  not  inferior 
and  holds  the  belief  that  nearly  half  our  logical  conclusions  rest 
upon  its  employment. 

In  the  above  case,  by  the  Law  of  Duality,  a  second  perpen- 
dicular can  or  can  -not  be  drawn.  It  was  shown  that  by  sup- 
posing that  a  second  one  could  be  drawn  led  us  to  an  absurdity. 
Hence,  a  second  can  not  be  drawn.  This  method  of  proof  is 
often  used  in  geometry. 

PROPOSITION  II. 

I.  Theorem.  If  two  adjacent  angles  have  their  exterior 
sides  in  a  straight  line,  these  angles  are  supplements 
of  each  other. 


SOPHUS  UK. 


393 


B 

\j 

II.     Given        the  exterior  sides  OA  and  OB  of  the  adjacent 
angles  AOD  and  BOD  respectively  and  the  straight 
AB  in  which  these  two  sides  lie. 
III.     To  prove  /  AOD  =  /  DOB. 

AOB  is  a  straight  line.     By  hypothesis. 
.'.  /  AOB  is  a  st.  /.     By  def.  of  a  st.  /. 
Kut/_AOD  + /_DOB  =  /_AOB.  By  Ax. 9. 
.*.  /'s  AOD  and  Z>&#  are  supplementary. 
By  def.  of  supl.  angles. 

Cor.  i.  The  sum  of  all  the  angles  about  a  point  in  a  plane 
is  equal  to  two  straight  angles. 

Cor.  2.  The  sum  of  all  the  angles  about  a  point  on  the  same 
side  of  a  straight  line  passing  through  a  point,  is  equal 
to  a  straight  angle: 

PROPOSITION  III. 

I.  Theorem.  CONVERSELY:  //  two  adjacent  angles  are 
supplements  of  each  other,  their  ez.  terior  angles  lie  in 
the  same  straight  line. 


B 


II.     Given         that  the  sum  of  the  adjacent  angles  AOD 
and  DOB  are  supplements  of  each  other,  that  is, 
equal  to  a  straight  angle. 
III.     To  prove  AO  and  OB  in  the  same  straight  line. 

1.  Assume  OF  in  the  same  straight  line  with 

OA. 

2.  Then  /_AOD  +  /_DOF\$z  straight  angle. 

By  Prop.  II. 

3.  But  /  AOD  +  /  DOB  is  a  straight  angle. 

By  hypothesis. 

4.  /.  /_AOD  +  /_DOB  =  /_AOD  +  /_DOF. 


IV.    Proof.  < 


By  Ax.  I. 

5.  £AOD=  /_AOD.     By  Law  of  Identity. 

6.  Subtracting  step  5  from  step  4,  /  DOB  - 

/  DOF.     By  Ax.  3. 


394 


FINKEL'S   SOLUTION   BOOK. 


7.  .'.    OB  and    OF  coincide.     By   converse 

Ax.  10. 

8.  .'.  AO  and  OB  are  in  the  same  straight 

line.     Q.  E.  D. 
SCHOLIUM.     Since  Propositions   II.  and   III.   are  true,  their 

opposites  are  true,  viz., 

//  the  exterior  sides  of  two  adjacent  angles  are  not  in  a  straight 
line,  these  angles  are  not  supplements  of  each  other. 

If  two  adjacent  angles  are  not  supplements  of  each  other,  their 
exterior  sides  are  not  in  the  same  straight  line. 


I. 


PROPOSITION  IV. 

Theorem.    If    one    straight    line    intersects    another 
straight  line,  the  vertical  angles  are  equal. 


IV.    Proof.  4 


II.     Given         the  two  lines  AB  and  DE  intersecting  in  0. 
III.     To  prove  /_  AOE  =  /_  DOB. 

1.     /  AOE  +  /_  AOD  equals   a   st.   /.By 
"  Prop.  I. 
/  AOD  +  /  DOB  equals   a  st.  /.     By 

Prop.  I. 
. • .  /  AOE  +  /  AOD  =  /  AOD  +  /  Z?O£. 

By  Ax.  1. 
Take  away  from  each  of  these  equals  the 

common  /  AOD. 

Then  /  AOE  =  /_  DOB.     By  Ax.  3. 
In  like  manner  we  may  prove  /  AOD  = 

/_  EOB.     Q.  E.  D. 
I.    Cor.    If  one  of  the  four  angles  formed  by  the  intersec- 
tion of  two  straight  lines  is  a  right  angle,  the  other  three 
angles  are  right  angles. 

PROPOSITION  V. 

I.     Theorem.    From  a  point  without  a  straight  line  one 
perpendicular,  and  only  one,  can  be  drawn  to  this  line. 


2. 


GEOMETRY. 


395 


IV.     Proof.  4 


II.     Given  the  point,  P,  and  the  straight  line,  AB. 

III.     To  prove  that  one  perpendicular  can  be  drawn  from 
P  to  AB,  and  only  one. 

1.  Turn  the  part  of  the  plane  above  AB  about 

AB  as  an  axis  until  it  falls  upon  the  part 
below  AB  and  denote  the  position  of  P 
by  Pr.  By  Assumption  3  of  the  Plane. 

2.  Turn  the  revolved  plane  about  AB  to  its 

original  position.  By  Assumption  3  of 
the  Plane. 

3.  Draw  the  straight  line  PP',  cutting  AB  in 

C.  By  Assumptions  1  and  2  of  the 
Straight  Line. 

4.  Take  any  other  point  D  in  AB,  and  draw 

PD  and  P'D. 

5.  Since  PCP'  is  a  straight  line,  PDP'  is  not 

a  straight  line. 

(Between  two  points  only  one  straight  line  can  be  drawn.) 

6.  Turn  the  figure  PCD  about  AB  until  P 

falls  on  P'.  By  Assumption  3  of  the 
Plane. 

7.  Then  CP  will  coincide  with  CP'  and  DP 

with  DP'. 

8.  /.  /  PCD  =   /_P'CD,  and  /  PDC  =  / 

P'DC.     Ax.  15. 

9.  .*.  /  PCD,  the  half  of  a  St.  /  PCP'  is  a 

right  /;  and  /  PDC,  the  half  of  /  PDP', 
is  not  a  right  angle. 

10.  .*.  PC  is  perpendicular  to  AB,  and  />/?  is 

not  perpendicular  to  AB.  By  def.  of 
Perpendicular. 

11.  .'.  one  perpendicular,  and  only  one,  can  be 

drawn  from  P  to  AB.     Q.  E.  D. 

PARALLEL  LINES. 

Definition.  Parallel  lines  are  lines  lying  in  the  same  plane 
and  never  meeting  however  far  produced. 

On  this  definition  and  the  assumption  of  parallel  lines  rests 
the  whole  theory  of  parallel  lines  in  Euclidean  geometry.  By 
convention,  we  say  that  parallel  lines  meet  at  infinity.  Why  this 
convention  is  adopted  will  become  apparent  in  studying  Higher 
Modern  Geometry. 

PROPOSITION  VI. 

I.     Theorem.      Two  straight  lines  in  the  same  plane  per- 
pendicular to  the  same  straight  line  are  parallel. 


396 


FINKEL'S  SOLUTION  BOOK. 


B 
D 


II.     Given        the  two  straight  lines  AB  and  CD  each  per- 
pendicular to  the  straight  line  AC. 

III.  To  prove  AB  and  CD  parallel. 

1.  AB  and  CD,  lying  in  the  same  plane,  must 

either  meet  or  not  meet.     By  Law  of 
Duality. 

2.  If  they  meet,  we  shall  have  two  lines  from 

the  same    point    perpendicular  to  the 
same  line.     By  hypothesis. 

IV.  PrOOf.  •{  (The  lines  AS  and  CD  being  perpendicular  to  AC.) 

But  this  is  impossible.     By  Prop.  V. 

(From  a  given  point  without  a  straight  line,  one  perpen- 
dicular, and  only  one,  can  be  drawn  to  a  straight  line.) 

.     .'.  AB  and  CD  cannot  meet,  however  far 

produced. 
.     .'.  AB  and  C£>are  parallel.     By  definition 

of  Parallel  Lines. 


PROPOSITION  VII. 

I.     Theorem.    If  a  straight  line  is  perpendicular  to  one  of 
two  parallel  lines,  it  is  perpendicular  to  the  other. 
H 


M  . 


c 


D. 


N 


K 


II. 


Given  the  parallel  lines  AB  and  CD  and  the  line 

//^perpendicular  to  AB. 

III.     To  prove    that  HK  is  perpendicular  to  CD. 

1.  Suppose  MN  drawn  through  ^perpendic- 

ular to  HK. 

2.  Then  MNis  parallel  to  AB.     By  Prop.  VI. 

(Two  lines  in  the  same  plane  perpendicular  to  the  same 
line  are  parallel.) 

^'     But  ^D  is  Parallel  to  AB.     By  hypothesis. 

4.  .  '  .     MN  coincides  with  CD.     By  assump- 

tion 1  of  parallel  lines. 

(Through  a  point  without  a  straight  line  only  one  straight 
line  can  be  drawn  parallel  to  that  line.) 

5.  CD  is  perpendicular  to  HK\  that  is, 
16.     HK'vs  perpendicular  to  CD.     Q.  E.  D. 


IV      Proof 


SIMON  NEWCOMB,  PH.  D.,  LL.  D. 


GEOMETRY.  397 


TRANSVERSALS. 

Definition.    A  straight  line  intersecting  two  or  more  straight 
lines  is  called  a  transversal. 


In  the  figure  EF  is  a  transversal  of  the  two  non-parallel  lines 
AB  and  CD. 

The  angles  AHI,  BHI,  CIH,  and  DIH  are  called  interior 
angles,  and  the  angles  AHE,  EHB,  GIF,  and  FID  are  called 
exterior  angles. 

The  angles  ^///  and  HID,  or  5J77  and  H/C  are  called  alter- 
nate-interior angles. 

The  angles  AHE  and  £>/F,  or  BHE  and  C/F  are  called  alter- 
nate-exterior angles. 

The  angles  AHE  and  C7tf,  ^///  and  CIFf  EHB  and  #/D,  or 
BHI  and  Z7/F  are  called  exterior-interior  angles. 

PROPOSITION  VIII. 

I.  Theorem.  If  two  parallel  lines  are  cut  by  a  third 
straight  line,  the  alternate-interior  angles  are  equal; 
and  conversely. 


I.    Find  the  value  of  an  angle  (1)  if  it  is  double  its  complement;  (2)  if 
it  is  one-fourth  of  its  complement. 

II.     Given     (1)  that  /  A  is  double  its  complement. 
III.     To  find  the  value  of  /  A. 

1.  rt.  /  — /  A  —  complement  of  /  A.  By  def. 

of  compl. 

2.  £A  =  2(rt.  /  —  /  A).     By  hypothesis. 

3.  .*.  /  A  =  2  rt.  /'s  —  2  /  A.     By  Distribu- 

tive Law  of  Multiplication. 

4.  Adding  2  /  A  to  these  two  equals,  we  have 

3  /  A  =  2  rt.  /'s.     By  Ax.  2. 

5.  .-.  £A  =  %rt.  /.     By  Ax.  7.     Q.  E.  F. 

Let  the  student  give  the  solution  of  (2). 

2.  Find  the  value  of  an  angle  (1)  if  it  is  three  times  its  supplement;  (2)  if  it  is 
one-third  of  its  supplement. 

3.  How  many  degrees  in  the  angle  formed  by  the  hands  of  a  clock  at  2  o'clock? 
3  o'clock?    4  o'clock?    9  o'clock? 


IV    So/nf  ion  <! 
m>  { 


398  FINKEL'S  SOLUTION  BOOK. 

PROPOSITION  IX. 

I.  Theorem.  If  two  parallel  lines  are  cut  by  a  third 
straight  Ime,  the  exterior  angles  are  equal,  and  con- 
versely. 

Let  the  student  give  the  demonstration  and  state  and  prove 
the  corollaries,  if  any,  coming  under  the  theorem. 


GEORGE  BRUCE  HAIySTED. 


GEOMETRY. 


GEOMETRY. 


1.      Geometry    is    the    science  that  treats  of   position    ana 
extension. 


2.     Pure  Geometry 


1.  Plane. 

2.  Solid. 


line  drawn  from 


Problem. — -To  bisect  a  given  triangle  by 
a  random  point  in  one  of  its  sides. 

Demonstration. — Let  ABC  be  the  given  triangle.  D  a  random 
point  in  the  side  BC,  and  E  the  middle 
point  of  BC.  Join  A  and  D,  A  and  E. 
Draw  EF  parallel  to  AD.  Draw  DF. 
Then  DF  bisects  the  triangle  ABC. 
For  the  triangle  ABE  is  equivalent  to 
the  triangle  AEC  (?).  The  triangle 
AFD  is  equivalent  to  the  triangle  FIG.  7. 

ADE  (?).     Hence,    ABDF  is    equivalent    to    ABE   (?)and, 
therefore,  DF  bisects  the  triangle^ BC.      Q.  E.  D. 

Proposition. —  The  square  described  upon  the  hypotenuse  of 
a  right  triangle  is  equal  to  theszim  of  the  squares  of  the  other  two 
sides. 

I.  Demonstration. — Let  CFD  be  any  right  triangle,  right  an- 
gled at  F  and  let  A  C,  CP,  and  DM 
be  the  squares  described  upon  its 
sides.  Then  the  square  A  C  is  equal 
to  the  sum  of  the  squares  CP  and 
DM.  Through  F,  draw  ^/^per- 
pendicular to  AB  and  produce  it  to 
meet  OP  produced,  in  G\  also  pro- 
duce BC  to  meet  OP  in  /and  AD 
to  meet  OP  produced,  in  R.  Draw 
GH  parallel  to  PD,  and  BT  par- 
allel to  CF.  Draw  AE.  Now  the 
triangles  COI  and  D  FC  are 
equal  (  ? ) .  Hence ,  C/=  CD=  CB, 
and  therefore  the  square  CJD=the 
parallelogram  CG  (?)=the  paral- 
lelogram BE  (?)=the  rectangle  FIG.  2. 
BK  (?).  In  like  manner,  the  square  DM  can  be  proved  equal 
to  the  rectangle  AK.  Hence,  the  square  A  C=the  square  CP-\- 
the  square  DM.  Q.  E.  D. 


400 


FINKEVS   SOLUTION   BOOK. 


II.  Demonstration. — Let  EDC\)Q  any  right  triangle,  right 
angled  at  D.  On  the  sides  DE  and 
DC  construct  the  squares  EDHG  and 
DCBM  respectively.  Produce  GE 
and  BC  until  they  meet  in  F,  form- 
ing the  square  F B A  G.  On  EC, 
the  hypotenuse,  construct  the  square 
ECKI.  Then  the  square  ECKI  is 
equal  to  the  sum  of  the  squares  EDHG 
and  D  CBM.  For,  the  square  GFBA 
is  equal  to  GEDH+DCBM+ 
2  EDCF  (=±ECF).  The  square 
GFBA  is  also  equal  to  the  square 
Hence,  ECKI+ 
GEDH-\-DCBM+±ECF  (?). 
=  GEDH+D  CBM.  Q.  E.  D. 


FIG.  .3 

Whence, 


ECKI 


Proposition. — In  any  triangle,  each  angle  formed  by  join- 
ing the  feet  of  the  perpendiculars  is  bisected  by  the  perpendicu- 
lar from  the  opposite  vertex. 


Demonstration. — Let  ABC  be  any  triangle  and  AD,  BE,  and 
CF  the  three  perpendiculars.  Join  D  and  E,  D  and  F,  and  E 
and  F. 

In  the  right  triangles  AEB  and  AFC,  the  angle  BA  C  is 
common  to  both.  Therefore,  they  are  similar.  Hence,  AB\A  C 
=AE\AF.  Now  the  triangles  BA  C  and  FAE  have  the  angle 
FAE  common  and  the  including  sides  proportional.  Therefore, 
they  are  similar,  and  the  angle  AFE=\\\e  angle  ACB.  In  a 
similar  manner  we  may  prove  that  the  angle  DFJ3=\he  angle 
A  CB  ;  the  angle  A  FE=the  angle  DFB. 
From  this  it  follows  that  the  angle  C  FA 
—the  angle  EFA=the  angle  CFB— 
the  angle  DFB.  Hence,  angle  EFC=snc\- 
gle  CFD  and  the  angle  EFD  is  bisected 
by  the  perpendicular  C  F.  In  a  similar 
manner, it  can  be  proved  that  A  D  bisects 
the  angle  FDE  and  EB  bisects  the  angle 
FED.  q.  E.  D.  FIG.  4. 

Problem. — From  a  given  point  in  an  arc  less  than  a  semi- 
circumference,  draw  a  chord  of  the  circle  which  will  be  bisected 
by  the  chord  of  the  given  arc. 

Demonstration. — Let  ABDC  be  the  given  circle,  AB  the 
given  arc,  AB  the  chord  of  the  arc,  and  P  any  point  of  the  arc 


PROFESvSOR  FELIX  KLEIN. 


GEOMETRY. 


401 


FIG.  5. 


A  PC.     Draw  the  diameter  POC  and  on  the  radius  PO  as  a  di- 

ameter describe  the  o\\c\?,PEO.  Then 

through  the  points^,  and  G,  of  inter- 

section draw  the  chords  PD  and  PF 

respectively,  and  they  will  be  bisected 

at  the  points  E  and    G.       For    draw 

DC    and    OE.      Then   the    triangles 

PJZOand  PD  Care  right  triangles(?) 

and  are  also  similar  (?).     Since  PEO 

and  PD  Care  similar,    the   line     OE 

is  parallel  to  DC,  and  since  O  is   the 

middle  point  of  P  C,E  is    the  middle 

point  of  P-D(  ?).     In  like   manner,  G 

is  the  middle  point  of  PF.          Q.E.  P 

Discussion.  —  There  are,  in  general,  two  solutions.  When  arc 
AB  is  diminished  until  B  coincides  with  A,  there  is  no  solution. 
When  AB  is  a  semi-circumference,  there  is  one  solution  and  the 
chord  is  the  diameter  POC. 

Proposition.  —  If  two  equal  straight  lines  intersect  each 
other  anywhere  at  right  angles,  the  quadrilateral  formed  by  join- 
ing their  extremities  is  equivalent  to  half  the  square  on  either 
straight  line. 

Demonstration.  —  Let  AB  and  CD  be  two  equal  straight  lines 
intersecting  each  other  at  right  angles 
at  E.  Join  their  extremities,  form- 
ing the  quadrilateral  A  CBD.  Then 
ACBD  is  equivalent  to  half  the 
square  of  AB  or  CD.  For  the  area 
of  the  triangle  A  CB  equals  \(AB 
X  CD)  and  the  area  of  the  triangle 
ADB  equals  ^(ABxED).  Hence, 
the  area  of  A 


FIG.  6. 


A  PROBLEM  IN  MODERN  GEOMETRY. 

An  equilateral  hyperbola  passes  through  the  middle  points 
D,  E,  and  F  of  the  sides  BC,  AC,  and  AB  of  the  triangle 
ABC,  and  cutting  those  sides  in  order  in  a,  ft,  and  y.  Show 
that  the  lines  Act,  Bft,  and  Cy  intersect  in  a  point  the  locus  of 
which  is  the  circumscribing  circle  of  the  triangle  ABC. 

Solution.  —  The  equation  to  any  conic  is  ua2  -\-vfi2  -\-ivy  '2  -f- 
Zu'fiy+Zv'ay+Zw'aft^O  ....  (1).  D  is  (0,  %a  sin  C,  \a  sin^)  ; 
E  (i^sinC,  0,  UsinA)  ;  F,  (-Jcsin  B,  ±cs\nA,0).  These 


by 


But   CD  equals  AB, 
hypothesis.       Hence,     ACBD 
Q.  E.  D. 


402  FINKEL'S   SOLUTION   BOOK. 

points  being  on  (1),  we  should  have  czv-{-b2TU-\-2bcu'=0  ...  (2), 
ctu+a^w+Zacv'^-O  .....  (3),  t>*t4+a2v  +  2al>iv'  =  0  ....(4). 

Whence  u=~(au'  —  bv/  —  CTV')  ....  (5),  v=  —  (bv'  —  civ'  —  au')  .  . 
0cv  •       ac^ 

.  .  .  (6),  w=—  (cwx  —  au'  —  bv')  ....  (7).  Substituting  in  the  con- 


dition  w-fv+w—  2z/  cosA—2v'  cos.#—  2w'cosC=0  .......  (8> 

that  (1)  is  an  equilateral  hyperbola, 


abc 

—1v'  cos  £—2<w'cosC—0  .....  (9).  Clearing  of  frac- 
tions and  noticing  that  2abc  cosA=a(t>2-{-c2  —  a2)  .....  (10), 
^abcco^=b(az+c^—  32)  ____  (11),  2a6ccos  C=c(a*+b2—  ^} 
.....  (12),  and  reducing,  «/cos^4-i;/cos^-)-w/cosC=0  ----  (13). 
Substituting  (5),  (6),  and  (7)  in  (1)  an  clearing  of  fractions, 


+y2+2u'a&c/3y+2v'a&cay-\-'2'w'aZ>ccx/3=Q  .  .  .  (14).    Where  this- 


cuts  BC,  a=0,  and  (14)  gives  &*(&/—  cw'—  au') 


=  —  CZ(CTV'  —  au'  —  bv')  .  .  .  (15),  whence  for  the  point  a;  al 

~  '    By  symmetry»  for  the  point. 


The 


is  found  to  be  £(  —  cwx+^vx  —  aw')/3  —  (cwx  —  bv'  —  au')y=0  . 
.  .  .  (18)  ;  to  Bft,  a(—c-w'+au'—bv')a—c(civ'—au'—bv')y=0. 
----  (19)  ;    and  to  Cy,  b(—au'+bv'—c>w'}p 
—  a(au'  —  bv'  —  civ')a=0  ....  (20)r  any  two  of   which    meet    in 


A,_ac  (  cw'—bv'—au'  )  (—c-w'+au'—bv'  ) 

r*  ~^r 

,     ab(—cw'+bv'—au')(—c'w'+au'—bv') 

~~DT 
The  circumscribing    circle   is  afly+bay-\-cctfi=O  ......  (22), 

which  is  satisfied  by  (21)  on  condition  (13).  proving  the  proposi- 
tion. 

NOTE.  —  This  problem  was  solved  by  Professor  William  Hoover,  A.  M., 
Ph.  D  ,  Professor  of  Mathematics  and  "Astronomy  in  the  Ohio  University, 
Athens,  Ohio,  who  is  one  of  the  leading  mathematicians  in  the  United 
States,  and  whose  biography  follows. 


GEOMETRY. 


BIOGRAPHY. 

PROF,  WILLIAM  HOOVER,  A.  M.,  PH.  D. 


Professor  Hoover  was  born  in  the  village  of  Smithville,  Wayne  county, 
Ohio,  October  17,  1850,  and  is  the  oldest  of  a  family  of  seven  children. 
Both  parents  are  living  in  the  village  where  he  was  born,  still  enjoying 
good  health. 

Up  to  the  age  of  fifteen  he  attended  the  public  schools,  and  for  two  or 
three  years  after,  a  local  academy.  Owing  to  needy  circumstances  he  was 
obliged  to  work  for  his  living  quite  early,  and  almost  permanently  closed 
attendance  at  any  kind  of  school  at  eighteen  years  of  age,  sometime  before 
which,  going  into  a  store  in  the  county  seat,  as  clerk.  Nothing  could  have 
been  farther  from  his  taste  than  this  work,  having  been  thoroughly  in  love 
with  study  and  books  long  before.  After  spending  two  or  three  years  in 
this  way,  he  went  to  teaching,  about  the  year  1869,  and  he  has  been  regularly 
engaged  in  his  favorite  profession  to  the  present  day. 

He  attended  Wittenberg  College  and  Oberlin  College  one  term  each,  a 
thing  having  very  little  bearing  on  his  education.  He  studied  no  mathe- 
matics at  either  place,  excepting  a  little  descriptive  astronomy  at  the  latter. 

After  teaching  three  winters  of  country  school,  with  indifferent  success, 
he  was  chosen,  in  1871,  a  teacher  in  the  Bellefontaine,  Ohio,  High  School, 
serving  one  year,  when  he  was  given  a  place  in  the  public  schools  of  South 
Bend,  Ind.  Remaining  there  two  years,  he  was  invited  to  return  to  Belle- 
fontaine as  superintendent  of  schools.  He  afterwards  served  in  the  same 
capacity  in  Wapakoneta,  O.,  two  years,  and  as  principal  of  the  second  dis- 
trict school  of  Dayton,  O.  In  1883,  he  was  elected  professor  of  mathemat- 
ics and  astronomy  in  the  Ohio  University,  Athens,  Ohio,  where  he  is  still 
in  service. 

Through  all  his  career  of  teaching,  Professor  Hoover  has  been  an  inces- 
sant student,  devoting  himself  largely  to  original  investigations  in  mathe- 
matics. Although  his  pretentions  in  other  lines  are  very  modest,  he  is  em- 
inently proficient  in  literature,  language,  and  history.  Before  going  into 
college  work  he  had  collected  a  good  library.  He  is  indebted  to  no  one  for 
any  attainments  made  in  the  more  advanced  of  these  lines,  but  by  indefati- 
gable energy  and  perseverance  he  has  made  himself  the  cultured,  classic,  and 
renowned  scholar  he  is. 

He  has  always  been  a  thorough  teacher,  aiming  to  lead  pupils  to  a  mas- 
tery of  subjects  under  consideration.  His  habits  of  mind  and  preparation 
for  the  work  show  him  specially  adapted  to  his  present  position,  where  he 
has  met  great  success.  He  studies  methods  of  teaching  mathematics, 
which  in  the  higher  parts  is  supposed  to  be  dry  and  uninteresting.  He  sets 
the  example  of  enthusiasm  as  a  teacher,  and  rarely  fails  to  impress  upon 
the  minds  of  his  students  the  immense  and  varied  applications  of  mathe- 
matics. He  is  kind  and  patient  in  the  class-room  and  is  held  in  the  highest 
esteem  by  his  students.  He  is  ever  ready  to  aid  the  patient  student  inquir- 
ing after  truth.  It  seems  to  be  a  characteristic  of  eminent  mathematicians 
that  they  desire  t'o  help  others  to  the  same  heights  to  which  they  them- 
selves haVe  climbed.  This  was  true  of  Prof.  Seitz;  it  is  true  of  Dr.  Martin; 
and  it  is  true  of  Prof.  Hoover. 

In  1879,  Wooster  University  conferred  upon  Prof.  Hoover  the  degree  of 
Master  of  Arts,  and,  in  1886,  the  degree  of  Doctor  of  Philosophy  cum  laude, 
he  submitting  a  thesis  on  Cometary  Perturbations.  In  1889,  he  was 
elected  a  member  of  the  London  Mathematical  Society  and  is  the  only 
man  in  his  state  enjoying  this  honor.  In  1890,  he  was  elected  a  member  of 
the  New  York  Mathematical  Society.  He  has  been  a  member  of  the  Asso- 


404  FINKEL'S   SOLUTION   BOOK. 

ciation  for  the  Advancement  of  Science  for  several  years.  Papers  accepted 
bv  the  association  at  the  meetings  at  Cleveland,  Ohio,  and  at  Washington, 
D.  C.,  have  been  presented  on  "The  Preliminary  Orbit  of  the  Ninth  Comet 
of  1886,"  and  "On  the  Mean  Logarithmic  Distance  of  Pairs  of  Points  in 
Two  Intersecting  Lines."  He  is  in  charge  of  the  correspondence  work  in 
mathematics  in  the  Chautauqua  College  of  Liberal  Arts  and  of  the  mathe- 
matical classes  in  the  summer  school  at  Lake  Chautauqua,  the  principal  of 
which  is  the  distinguished  Dr.  William  R.  Harper,  president  of  the  new 
Chicago  University.  The  selection  of  Professor  Hoover  for  this  latter  po- 
sition is  of  the  greatest  credit,  as  his  work  is  brought  into  comparison  with 
some  of  the  best  done  anywhere. 

He  is  a  critical  readei  and  student  of  the  best  American  and  European 
writers,  and  besides,  is  a  frequent  contributor  to  various  mathematical  jour- 
nals, the  principal  of  which  are  School  Visitor,  Mathematical  Messenger, 
Mathematical  Magazine,  Mathematical  Visitor,  Analyst,  Annals  of  Math- 
ematics, American  Mathematical  Monthly,  and  Educational  Times,  of  Lon- 
don, England. 

His  sU'le  is  concise  and  his  aim  is  elegance  in  form  of  expression  of 
mathematical  thought.  While  greatly  interested  in  the  various  branches  of 
pure  mathematics,  he  is  specially  interested  in  the  applications  to  the  ad- 
vanced departments  of  Astronomy,  Mechanics,  and  the  Physical  Sciences 
— such  as  Heat,  Optics,  Electricity,  and  Magnetism.  The  "electives"  of- 
fered in  the  advanced  work  for  students  in  his  University  are  among  the 
best  mathematics  pursued  an>  where  in  this  country. 

He  is  an  active  member  of  the  Presbyterian  church  and  greatly  interested 
in  every  branch  of  church  work.  He  has  been  an  elder  for  a  number  of 
years  and  was  chosen  a  delegate  to  the  General  Assembly  meeting  at  Port- 
land, Oregon,  in  May,  1892,  serving  the  church  in  this  capacity  with  fidel- 
ity and  intelligence.  In  this  biography  of  Professor  Hoover,  there  is  a  val- 
uable lesson  to  be  learned.  It  is  this:  Energy  and  perseverance  will  bring 
a  sure  reward  to  earnest  effort.  We  see  how  the  clerk  in  a  county  seat 
store,  in  embarrassing  circumstances  and  unknown  to  the  world  of  thinkers, 
became  the  well  known  Professor  of  Mathematics  and  Astronomy  in  one  of 
the  leading  Institutions  of  learning  in  the  State  of  Ohio.  "Not  to  know 
him  argues  yourself  unknown." 


THE  NINE-POINT  CIRCLE. 

Proposition. — If  a  circle  be  described  about  the  pedal  trian- 
gle of  any  triangle,  it  will  pass  through  the  middle  points  of  the  lines 
drawn  from  the  orthocenter  to  the  vertices  of  the  triangle,  and 
through  the  middle  points  of  the  sides  of  the  triangle,  in  all,  through 
nine  points.  (2)  The  center  of  the  nine-point  circle  is  the  mid- 
dle point  of  the  line  joining  the  orthocenter  and  the  center  of  the 
circumscribing  circle  of  the  triangle.  (8)  The  radius  of  the  nine- 
point  circle  is  half  the  radius  of  the  circumscribing  circle  of  the 
triangle.  (4)  The  centroid  of  the  triangle  also  lies  on  the  line 
joining  the  orthocenter  and  the  center  of  the  circumscribing  circle 
of  the  triangle,  and  divides  it  in  the  ratio  of  2\  1<  (5)  The  sides  of 
the  pedal  triangle  intersect  the  sides  of  the  given  triangle  in  the 
radical  axis  of  the  circumscribing  and  nine-point  circles.  (6)  The 
nine-point  circle  touches  the  inscribed  and  escribed  circles  of  the 
triangle. 


BENJAMIN  PEIRCE. 


GEOMETRY.  405 

The  Pedal  Triangle  is  a  triangle  formed  by  joining  the 
feet  of  the  perpendiculars  drawn  from  the  vertices  of  a  triangle 
to  the  opposite  sides. 

The  Orthocenter  is  the  point  of  intersection  of  these  per- 
pendiculars. 

Medial  Lines,  or  Medians,  are  lines  drawn  from  the 
vertices  of  a  triangle  to  the  middle  point  of  the  opposite  sides. 

The  Centroid    is  the  point  of  intersection  of  the  medians. 

The  Radical  Axis  of  two  circles  is  the  locus  of  the  points 
whose  powers  with  respect  to  the  two  circles  are  equal. 

Demonstration. — Let^47?Cbe  an)  triangle,  AD,  BF,  and 
CE  the  perpendiculars  from  the  vertices  to  the  opposite  sides  of 
the  triangle.  O  is  the.  orthocenter.  Join  the  points  F,  E,  and 
D.  Then  FED  is  the  pedal  triangle.  About  this  triangle,  de- 
scribe the  circle  FEHDK.  It  will  then  pass  through  the  mid- 
dle points  Z,  N)  and  R  of  the  lines,  OA,  OB,  and  OC,  and  the 
middle  points  H,  G,  and  A" of  the  sides  AB,  BC,  and  A  C,  in  all, 
through  nine  points. 

Since  the  angles  AFO  and  AEO  of  the  quadrilateral  are  both 
right  angles  a  circle  may  be  described  about  it.  For  the  same 
reason  circles  may  be  described  about  the  quadrilaterals  EBDO 
andODCF.  Draw  the  lines  7^7?  and  RG.  Now  the  angles 
FRE  and  FDE  are  equal,  being  measured  by  half  the  same 
arc  FE.  But  FDE  equals  1EDL,  because  AD  bisects  the 
angle  EDF.  .'.  FRO  equals  %FDL.  Both  being  measured  by 
the  same  arc  OF,  and  FRO  being  two  times  FDL,  FRO  is  an 
angle  at  the  center;  therefore,  since  OC  is  the  diameter  of  the 
circle  circumscribing  FODC,  R  is  the  middle  point  of  OC.  In 
like  manner  it  may  be  proved  that  OB  and  OA  are  bisected  in 
the  points  TV  and  JL  respectively.  Draw  the  line  RG.  The 
angles  RGC  and  RGB  are  equal  to  two  right  angles.  Also  the 
angles  R  GB  and  RED  are  equal  to  two  right  angles,  because 
they  are  opposite  angles  of  a  quadrilateral  inscribed  in  a  circle. 
Therefore  R  GC  is  equal  to  RED.  But  RED  is  equal  to 
OBD,  because  both  are  measured  by  half  the  arc  OD.  .-.  The 
angle  RGC  equals  the  angle  OBD,  and  consequently  the  line 
RG  is  parallel  to  the  line  OB.  But,  since  RG  bisects  OC  in  R 
and  is  parallel  to  OB,  it  bisects  B  C  in  G.  In  like  manner,  it  may 
be  shown  that  AB  and  A  C  are  bisected  by  the  nine-point  circle 
in  the  points  H and  A" respectively.  Hence,  the  circle  passes,  in 
all,  through  nine  points.  Q.  E.  D. 

(2.)  Draw  the  perpendiculars GP,  KP,  and  HP  from  the 
-middle  points  of  the  sides  of  the  triangle.  They  all  meet  in  a 
common  point  P  which  is  the  center  of  the  circumscribing  circle 
•of  the  triangle.  With  P  as  a  center  and  radius  equal  to  PB, 


406 


FINKEL'S  SOLUTION  BOOK 


describe  the  circumscribing  circle.     Draw  the  perpendiculars  SY~ 
SJ,  and  SZ  to  the  middle  points   of  the   chords   FK,  EH,  and 


FIG.  6. 

DG.  These  all  meet  in  the  same  point  S,  which  is  the  center  of 
the  nine-point  circle.  In  the  trapezoid  PHEO,  since  SJ  bi- 
sects EH  in  J  and  is  parallel  to  PH,  it  bisects  OP  in  S.  Hence, 
the  center  of  the  nine-point  circle  is  the  middle  point  of  the  line 
joining  the  orthocenter  and  center  of  the  circumscribing  circle. 

Q.  E.  D. 

(3.)  Draw  the  lines  KN  and  PB.  Since  the  angle  KFN  is 
a  right  angle,,  the  line  KN  is  a  diameter  of  the  nine-point  cir- 
cle. KP=%BO=BN.  Since  KP  and  BN  are  equal  and 
parallel,  KPBN  is  a  parallelogram,  and  consequently  KN—BP. 
.-.  SN—\fiP.  But  SJV  is  the  radius  of  the  nine-point  circle 
and  BP  is  the  radius  of  the  circumscribing  circle  of  the  triangle. 
Hence,  the  radius  of  the  nine-point  circle  is  half  the  radius  of 
the  circumscribing  circle.  Q.  E,  D. 

(4.)  Draw  the  medial  lines  BK,  AG,  and  CH.  Draw  the 
line  KH.  Now  the  triangles  KPH  a\\&  BOC  are  similai  he- 


GEOMETRY. 


40T 


cause  the  sides  of  the  one  are  respectively  parallel  to  the  sides  of 
the  other,  and  the  line  UK  is  half  the  line  BC,  because  //and 
Jfare  the  middle  points  of  the  sides  AB  and  A  C.  .'.BO=ZKP. 
The  triangles  KPQ  and  BOQ  are  similar,  because  the  angles  of 
one  are  respectively  equal  to  the  angles  of  the  other.  Then  we  have 
KP:KQ\:BO:BQ  or  KP'.BO:  :KQ  :  BQ.  But  BO=>IKP. 
.-.  BQ=ZKQ.  .'.  Q  is  the  centroid  and  divides  the  line  joining 
orthocenter  and  the  center  of  the  circumscribing  circle  in  the 
ratio  of  2:1.  Q.E.  D. 

Hence  the  line  joining  the  centers  of  the  circumscribing  and 
nine-point  circles  is  divided  harmonically  in  the  ratio  of  2  :  1  by 
the  centroid  and  orthocenter  of  the  triangles.  These  two  points 
are  therefore  centers  of  similitude  of  the  circumscribing  and 
nine-point  circles.  .-.  Any  line  drawn  through  either  of  these 
points  is  divided  by  the  circumferences  in  the  ratio  of  2  :  1. 

(5.  )  Produce  FE  till  it  meets  B  C  in  P/.  Since  two  opposite  an- 
gles of  the  quadrilateral  BEFC  are  equal  to  two  right  angles,  a 
circle  may  be  circumscribed  about  it.  Then  we  have  P/  E.  P/  F 
=P'B.  P'  C  ;  therefore  the  tangents  from  P'  to  the  circles  are 
equal.  Q.  E.  D. 

(6.)  Let  Obe  the  orthocenter,  and  /and  (Hhe  centers  of  the  in- 
scribed and  circumscribed  circles.  Produce  AI  to  bisect  the  arc 
BC'in  T.  Bisect  AO  in  L,  and 
join  GL,  cutting  A  Tin  S.  The 
nine-point  circle  passes  through 
Gj  D,  and  L,  and  D  is  a  right  an- 
gle. Hence,  GL  is  a  diameter, 
and  is  therefore  =R=QA.  There- 
fore GL  and  QA  are  parallel.  But 
QA=QT,  therefore  GS=GT= 
CTsm$A=2fisin^A.  Also  ST 
=2GScos8,  6  being  the  angle  GST 
—  GTS.  N  being  the  center  of 
the  nine-point  circle,  its  radius 
—NG=%R  ;  and  r  being  the  radius 
of  the  inscribed  circle,  it  is  required 
to  show  that  NI=NG—r.  NI* 
=SN*  -f-  5/2—2  SN.  S/cos  6.  Sub- 
stitute SN=±R—  GS;  SI=TI— 
ST=2fism%A—2GScos0;  and  G 
proposition.  If  J  be  the  center  of  the  escribed  circle  touch 
ing  B  C,  rl  its  radius,  it  is  shown  in  a  similar  way  that 
JVJ=NG+rl. 


FIG.  7. 

to   prove    the 


408  FINKEIv'S  SOLUTION   BOOK. 

THE  THREE  FAMOUS  GEOMETRICAL  PROBLEMS  OF 

ANTIQUITY. 

The  limits  of  this  work  forbid  our  carrying  the  discussion  of 
elementary  geometry  further.  We  have  given  merely  an  outline 
of  how  the  subject  may  be  studied  by  the  student  and  presented 
by  the  teacher  and  that  is  our  chief  aim  in  this  work.  But  before 
leaving  the  subject,  it  will  be  of  interest  to  briefly  speak  of  three 
famous  problems  in  geometry,  —  problems  that  have  profoundly 
interested  the  mathematicians  from  the  time  of  Plato  down  to 
the  present  time.  These  problems  have  been  referred  to  before 
in  this  book  so  that,  at  this  point,  we  shall  only  bring  them  to- 
gether and  speak  of  them  more  explicitly.  The  problems  re- 
ferred to  are, 

(i)  The  Duplication  of  the  Cube;  (ii)  the  Trisection 
of  an  Angle;  (iii)  the  Quadrature  of  the  Circle. 

The  first  of  these  problems  means  to  find  the  edge  of  a  cube 
whose  volume  shall  be  twice  that  of  a  given  cube;  the  second 
means  to  divide  any  given  angle  into  three  equal  parts;  and 
the  third  means  to  find  the  side  of  a  square  whose  area  shall  be 
equal  to  that  of  a  given  circle. 

As  has  been  said,  constructions  in  pure  geometry  or  Euclidean 
Geometry  admit  of  the  use  of  an  ungraduated  ruler  and  a  pair 
of  compasses  only.  With  this  restriction,  all  three  problems 
are  insoluble.  This  is  the  important  point  to  be  observed.  The 
problems  are  only  impossible,  because  we  are  limited  in  the  use 
•of  our  instruments  to  a  straight  edge,  or  ungraduated  ruler,  and 
a  pair  of  compasses.  In  this  way  many  problems  may  be  made 
impossible.  For  example,  it  is  impossible  to  go  across  the  At- 
lantic Ocean  from  Boston  to  Liverpool  on  a  bicycle,  but  with  a 
steamship  the  trip  is  made  very  easily.  So  too,  if  other  instru- 
ments are  used  our  three  problems  are  easily  solved.  The  solu- 
tions of  the  first  and  second  problems  are  implicitly  involved  in 
the  Galois  theory  as  presented  to-day  in  treatises  on  higher 
algebra.  The  impossibility  of  the  solution  of  the  third  was 
demonstrated  in  1882  by  Lindemann. 

The  first  two  problems  may  be  reduced  to  one,  viz.,  that  of 
finding  two  means  between  two  given  extremes.  In  the  first  prob- 
lem, if  we  let  a  be  the  edge  of  given  cube  and  x  that  of  the  re- 

?uired  cube,  then  we  must  have  x*  =  2a3,  or  a  :  JT  =  x  :  y  =  y  :  20. 
n  the  second,  if  a  is  the  sine  of  the  given  angle,  and  x  the  sine 

of  one-third  the  angle  ;  then  4^  =  $x  —  a,  or  I  :  $x  =  $x  '•  y  — 


The  problem  of  the  duplication  of  the  cube  was  known  in 
ancient  times  as  the  Delian  problem,  in  consequence  of  a  legend 
that  the  Delians  had  consulted  Plato  on  the  subject.  It  is  as- 
serted by  Philoponus,  that  the  Athenians  in  430  B.  C.  were  suf- 


PROBLEMS. 


409- 


fering  from  the  plague  of  eruptive  typhoid  fever  and  in  order  to 
stop  it  consulted  the  oracle  at  Delos  as  to  how  it  might  be  done. 
Appolo  replied  that  they  must  double  the  size  of  the  altar  of 
Minerva  which  was  in  the  form  of  a  cube.  This  to  the  un-' 
learned  suppliant,  was  an  easy  task,  and  a  new  altar  having 
each  of  its  edges  double  that  of  the  old  one  was  constructed,, 
in  consequence  of  which  the  volume  was  increased  eight-fold. 
This  so  enraged  the  god  that  he  made  the  pestilence  worse  than 
before,  and  informed  a  fresh  deputation  that  it  was  useless  to 
trifle  with  him  as  the  new  alter  must  be  a  cube  and  have  a. 
volume  exactly  double  that  of  the  old  one.  Suspecting  a  mystery,, 
the  Athenians  applied  to  Plato  who  referred  them  to  the  geom- 
etricians. In  an  Arab  work,  it  is  related  that  Plato  replied  to 
them,  saying,  Ye  have  been  neglectful  of  the  science  of  geometry 
and,  therefore,  hath  God  chastised  you,  since  geometry  is  the 
most  sublime  of  all  the  sciences. 

Many  solutions  of  this  problem  have  been  given,  one  of  which 
is  given  on  page  234,  by  means  of  the  Cissoid.  We  here  give 
another  by  means  of  the  parabola. 


Let y*  —  ax,  be  the  equation  of  the  parabola  whose  axis  coin- 
cides with  axis  of  abscissas  and  x*  —  2#j/,  the  equation  of  the 
parabola  whose  axis  coincides  with  the  axis  of  ordinates.  Solv- 
ing these  two  equations,  we  find  j3  —  2#3,  that  is,  PMS  =  2a3. 
Hence,  if  a  is  the  edge  of  the  given  cube  PMis  the  edge  of  the 
required  cube. 

To  trisect  an  angle,  we  proceed  as  follows: 

Let  AOB  be  the  given  angle.  With  O  as  center  and  any 
radius,  describe  a  circle,  ABD.  Draw  the  secant  BDCso  that 
DC  shall  be  equal  to  the  radius  OB.  (This  is  impossible  unless 
a  graduated  ruler  is  used.)  Then  draw  OD.  Then  angle  BC^ 
=£  angle  AOB.  For  angle  DCO  =  angle  DOC.  Why?  Angle 
BDO  =  2  angle  DCO.  Why?  Angle  DCO  +  angle  CBO  = 
angle  BOA.  Why?  .'.  Angle  AOB  =  3  angle  BCO.  Why? 


410  FINKEL'S  SOLUTION  BOOK. 

The  following  elegant  solution  is  due  Claraut 


Let  AOB  be  the  given  angle.  With  O  as  center  and  any 
radius  describe  a  circle.  Draw  AB  and  trisect  it  in  //"and  K,  so 
that  AH=  HK  =  KB.  Bisect  the  angle  AOB  by  OCt  cutting 
AB  in  L.  Then  AH  =  ^HL.  With  focus  A,  vertex  H,  and 
directrix  OC,  describe  a  hyperbola.  Let  the  branch  of  this 
hyperbola  which  passes  through  H  cut  the  circle  in  P.  Draw 
PM  perpendicular  to  OC  and  produce  it  tq  cut  the  circle  in  Q. 
Then  by  the  focus  and  directrix  property,  we  have  AP :  PM  = 
AH:HL  =  2:\.  .'.AP=2  PM=  PQ.  Hence,  by  symmetry, 
AP  =  PQ  =  QR.  Hence,  AOP  =  POQ  =  QOR. 

The  Quadrature  of  the  Circle  is  effected  by  the  Quadratrix. 
See  page  238. 

For  a  very  full  treatment  of  these  problems,  see  Klein's 
Famous  Problems  of  Elementary  Geometry,  translated  from  the 
German  by  Professors  Beman  and  Smith,  also  see  Mathematical 
Recreations  and  Problems,  by  W.  W.  R.  Ball. 


JAMES  JOSEPH  SYLVESTER,  U,.  D.,  F.  R.  S. 


PROPOSITIONS.  411 

PROPOSITIONS. 

1.  The  lines  which  join  the  middle  points  of  adjacent  sides  of  any  quad- 
rilateral, form  a  parallelogram. 

2.  Two  medians  of  a  triangle  are  equal;  prove   (without  assuming    that 
they  trisect  each  other)  that  the  triangle  isisoscles. 

3.  In  an  indefinite  straight  line  AB  find  a  point  equally  distant  from 
two  given  points  which  are  not  both  on  AB. 

When  does  this  problem  not  admit  of  solution? 
Construct  a  right  triangle  having  given: 

4.  The  hypotenuse  and  the  difference  of  the  sides. 

5.  The  perimeter  and  an  acute  angle. 

6.  The  difference  of  the  sides  and  an  acute  angle. 

7.  Construct  a  triangle  having  given  the  medians. 

8.  Construct  a  triangle,  having  given  the  base,  the  vertical  angle,  and(l) 
the  sum  or  (2)  the  difference  of  the  sides. 

9.  Describe  a  circle  which  shall  touch  a  given  circle  at  a  given  point,  and 
also  touch  a  given  straight  line. 

10.  Describe  a  circle  which  shall  pass  through  two  given   points  and 
be  tangent  to  a  given  line. 

11.  Find  the    point  inside   a  given  triangle  at  which    the  sides  subtend 
equal  angles. 

12.  Describe  a  circle  which  shall  be  tangent  to  two  intersecting  straight 
lines  and  passing  through  a  given  point. 

13.  Divide  a  triangle  in  two  equal  parts  by  a  line  perpendicular  to  a 
side. 

14.  Inscribe  in  a  triangle,  a  rectangle  similar  to  a  given  rectangle. 

15.  Construct  an  equilateral  triangle  equivalent  to  a  given  square. 

16-  Trisect  a  triangle  by  straight  lines  drawn  from  a  given  point  in  one 
of  its  sides 

17.  Draw  through  a  given  point  a  straight  line,  so  that  the  part  of  it  in- 
tercepted between  a  given  straight  line  and  a  given    circle  may  be  divided 
at  the  given  point  in  a  given  ratio. 

18.  Construct  a  circle  equivalent  to  the  sum  of  three  given  circles. 

19.  Find  the  locus  of  a  point  such  that  the  sum  of  its  distances  from  three 
given  planes  is  equal  to  a  given  straight  line. 

20.  Construct    a    sphere    tangent   to    three   given  spheres    and    passing 
througn  a  given  point. 

21.  Draw  a  circle  tangent  to  three  given  circles. 

NOTE. — This  proposition  is  known  as  the  Taction  Problem,  It  was  pro- 
posed and  solved  by  Apollonius,  of  Pergse,  A.  D.  200.  His  solution  was  indi- 
rect, reducing  the  problem  to  ever  simpler  and  simpler  problems.  It  was 
lost  for  centuries,  but  was  restored  by  Vieta.  The  first  direct  solution  was 
given  by  Gergonne,  1813.  An  elegant  solution  of  this  problem  is  given  by 
Prof.  E.  B.  Seitz,  School  Visitor,  Vol.  IV,  p.  61. 

22.  Construct  a  sphere  tangent  to  four  given  spheres. 
NOTE.— This  problem  was  first  solved  by  Fermat  (1601—1665). 

23.  The  perpendicular  from  the  center  of  gravity  of  a  tetrahedron  to  any 
plane  without  the  tetrahedron  is  one-fourth  of  the   sum    of  the  perpendic- 
nlarsfrom  the  vertices  to  the  same  plane. 


412  FINKEL'S    SOLUTION    BOOK. 

1.  Define:    a  segment  of  a  circle,  four  proportional  magnitudes,  two 
similar  polygons,  the  projection  of  a  segment  of  a  straight  line  on  another 
straight  line. 

2.  The  sum  of  all  the  plane  angles  about  a  point 'is  four  right  angles. 

3.  The  locus  of  all  points  equally  distant  from  two  fixed  points  is  the 
s.raight  line  that  bisects  the  line  joining  the  two  points,  at  right  angles. 

4.  A  straight  line  that  is  perpendicular  to  a  radius  at  its  extremity  is. 
tangent  to  the  circle ;    and  conversely. 

5.  Two  polygons  that  are  similar  to  a  third  polygon  aie  similar  to- 
each  other. 

6.  If  two  triangles  have  an  angle  of  the  one  equal  to  an  angle  of  the 
other,  their  areas  are  to  each  other  as  the  rectangles  of  the  sides  includ- 
ing those  angles. 

7.  The  ratio  of  the  circumference  of  a  circle  to  its  diameter  is  the 
same  for  all  circles. 

8.  Find  the  side  of  the  largest  square  that  can  be  cut  from  a  tree  whose 
circumference  is  14  feet. 

Cornell  University  —  Entrance  Examination,  1899. 

[Proofs  by  limits  are  not,  in  general,  satisfactory.] 

1.  Define :    a  plane,  a  straight  line  perpendicular  to  a  plane,  a  straight 
line  parallel  to  a  plane,  two  parallel  planes,  a  diedral  angle,   the  plane 
angle  of  a  diedral. 

2.  The  sum  of  the  face  angles  of  a  convex  polyderal  angle  is  less 
than  four  right  angles. 

3.  The  sections  of  a  prismatic  surface  made  by  two  parallel  planes  are 
equal  polygons. 

4.  The  frustum  of  a  triangular  pyramid  is  equal  in  volume  to  three 
pyramids,   whose   common  altitude  is   the  altitude   of  the   frustum,   and 
whose  bases  are  the  two  bases  of  the  frustum  and  a  mean  proportional 
between  them. 

5.  To  draw  a  plane  tangent  to  a  cylindar  with  circle  base. 

6.  If  two  angles  of  a  spherical  triangle  be  equal,  the  opposite  sides 
are  equal. 

7.  The  lateral  area  of  a  cone  of  revolution  is  half  the  product  of  the 
perimeter  of  its  base  and  its  slant  height. 

8.  A  cylindrical  pail  is  6  inches  deep  and  7  inches  in  diameter :    find 
how  much  water  it  holds,  and  how  much  tin  it  takes  to  make  it. 

Cornell  University  —  Entrance  Examination, 


1.  Define  a  straight  line   (preferably  without  using  the  ideas  of  dis- 
tance  or   direction).     Also   define:     equal,    greater,   limit   of   a   variable, 
length  of  a  curve. 

2.  If  two  triangles  have  two  sides  of  one  equal  to  two  sides  of  the 
other,    and   the    included   angles   of   the   first   greater    than    that   of   the 
second,   prove  that  the  third   side  of  the  first  is  greater  than  the  third 
side  of  the  second.     Also  prove  the  converse  of  this  theorem. 

3.  Similar  triangles  (and  similar  polygons)  are  to  each  other  as  the 
squares  on  homologous  sides. 

4.  Construct  a   triangle,   being  given   the  lengths   of  the   three  per- 
dendiculars  from  the  vertices  on  the  opposite  sides. 

5.  Compute   the   side   of   a   regular   pentagon   inscribed   in   a   circle 
whose  radius  is  given. 


PROBLEMS.  413 

6.  Two  straight  lines  in  space  have  one  and  but  one  common  per- 
pendicular, and  it  is  the  shortest  line  that  can  be  drawn  from  one  to 
the  other. 

7.  Compute  the  volume  of  a  regular  octahedron  whose  edge  is  two 
units. 

8.  Show  how  to  find  the  radius  of  a  given  sphere  by  means  of  meas- 
urements made  on  the  surface. 

9.  Prove  that  the  volume  of  a  cone  is  equal  to  the  area  of  the  base 
multiplied   by  one-third   of  the   altitude.     Also   state   without   proof   the 
chain  of  propositions  which  lead  up  to  this  theorem. 

10.  Find  the  locus  of  a  point  in  space  the  ratio  of  whose  distances 
from  two  given  points  is  equal  to  a  given  constant. 

Cornell  University  —  Scholarship  Examination,  1899, 
Time,  3  hours. 

1.  State  and  prove  the  theorem  of  Menelaus  —  and  its  inverse. 

2.  Prove:     Circles  described  on  any  three  chords  from  one  point  of 
a  circle  as  diameters,   have  their  other  three  points  of  intersection  co- 
straight. 

3.  Explain  the  Peaucellier  Cell. 

4.  State  and  prove  the  dual  theorem  of:     The  pole  of  any  straight 
through  a  point  is  on  the  polar  of  the  point. 

5.  Prove:    The  diagonal  triangle  of  a  cyclic  quadrangle  is  self-conju- 
gate (its  own  reciprocal  polar). 

6.  (a)     Explain  what  is  meant  by  a  cross  ratio  of  a  range  of  four 
points. 

(fr)  If  (PQRS)  =3,  what  are  the  other  distinct  cross  ratios  of  the 
same  range,  and  what  are  their  magnitudes? 

(c)  Deduce  the  distinct  values  of  the  cross  ratios  of  a  harmonic 
range. 

7.  Prove  Pascal's  theorem  concerning  a  cyclic  hexagon. 

(£)     State  the  corollaries  as  the  number  of  sides  is  diminished. 

8.  What  is  the  radical  axis  of  two  circles? 

(b)     Prove:     The  difference  between  the  squares  of  the  tangents 

from  any  point  to  two  circles  equals  twice  the  rectangle  of  the  center  sect 

of  the  circles  and  the  perpendicular  sect  from  the  point  to  the  radical  axis. 

Examination  in  Halsted's  Modern  Geometry.     The  University 

of  Texas,  1894.     Time,  3  hours. 

1.  From  the  common  notion  "solid"  as  a  starting  point,  define  surface, 
line,  point,  plane,  straight  line. 

(b)  Define  angle,  and  point  out  the  angles  determined  by  a  bi- 
radial. 

(r)  What  is  meant  by  the  statement  that  two  magnitudes  are 
equivalent?  —  that  one  magnitude  is  greater  than  another? 

(d)  Define  the  terms,  multiple,  submultiple,  fraction,  ratio. 

(e)  What  is  the  direct  meaning  of  the  statement  that  two  series 
of  magnitudes  are  proportional?    State  the  simplest  criteria  of  propor- 
tionality between  two  series  of  magnitudes  in  which  to  every  one  of  either 
series  there  corresponds  one  of  the  other  series.     Apply  the  test  to  two 
such  series  where  it  is  fulfilled ;   and  again  where  one  criterion  fails. 

(/)     When  are  two  figures  perspective?  —  when  similar? 

2.  Discuss  the  problem:     To  describe  a  circle  tangent  to  three  given 
intersecting  straights,  not  all  through  the  same  point.  —  Also,  the  analogous 
problem  in  a  sphere. 


414  FINKEL'S    SOLUTION    BOOK. 

3.  (a)     State  the  conditions  of  congruence  of  two  plane  triangles. 

(b)  State  the  conditions  of  similarity  of  two  plane  triangles. 

(c)  State  the  conditions  of  congruence  of  two  spherical  triangles. 

4.  (a)     Investigate  the  form  of  the  quadrilateral  made  by  joining  the 
mid  points  of  consecutive  sides  of  a  quadrilateral. 

(fr)     Its  relative  size. 

5.  Divide  a  sect  internally  and  externally  in  a  given  ratio. 

6.  (a)     Prove:     If  four  sects  are  proportional  the  rectangle  of  the 
extremes  is  equivalent  to  the  rectangle  of  the  means. 

(b)     State  the  inverse. 

7.  (a)     Prove  the  Pythagorean  theorem  without  using  any  other  con- 
cerning equivalence  of  figures. 

(b)     Prove:     The  altitude  to  the  hypothenuse  is  a  mean  propor- 
tional between  the  segments  of  the  hypothenuse. 

8.  (a)     When  is  one  spherical  triangle  A'B'C  the  polar  of  another. 
ABC? 

(fr)     Prove:     The  sides  of  a  spherical  triangle  intersect  the  cor- 
responding sides  of  its  polar  on  the  polar  of  their  orthocenter. 

Examination  in  Hoisted' s  Elementary  Synthetic  Geometry.     The 
University  of  Texas,  1894.     Time,  3  hours. 


ALGEBRA.  415 


ALGEBRA 


1 .  Algebra  is  that  branch  of  mathematics  which  treats  of 
the  general  theory  of  operations  with  numbers,  or  quantities. 

The  operations  of  ordinary  abstract  arithmetic  are  a  particular  case  of 
algebra.  Thus,  algebra  is  sometimes  called  generalized  arithmetic  and  in 
turn  arithmetic  is  sometimes  called  specialized  algebra. 

2.  An  Operation,  in  mathematics,  is  the  act  of  passing 
from  one  number  to  another,  the  second  number  having  a  defi- 
nite relation  to  the  first. 

3.  An  Operator,  in  mathematics,  is  a  letter  or  symbol 
designating  the  operation  to  be  performed. 

Thus,  -{-,  — ,  X»  -*-»  Vi  — »  or  &  x  >  and  f  are  operators. 
dx  «l 

4.  The  Fundamental  Laws  of  Algebra  are  the  Com- 
mutative Law;  the  Associative  I/aw;  and  the  Distribu- 
tive L/aw. 

The  Commutative  L/aw  states,  that  additions  and  subtrac- 
tions may  be  performed  in  any  order ;  also  the  factors  of  a  product 
may  be  taken  in  any  order. 

The  Associative  If  aw  states,  that  the  terms  of  an  expres- 
sion may  be  grouped  in  any  order.  Thus,  a+6— c+d—  e=(a+b)— 
c+(d-e)=a+(b-c)+(d-e)=a+b-(c—d}-e. 

Also  the  factors  of  a  product  may  be  grouped  in  any  order. 

The  Distributive  I/aw  states,  that  the  product  of  a  com- 
pound expression  by  a  single  factor  is  the  algebraic  sum  of  the 
partial  products  of  each  term  of  the  compound  expression  by  that 
factor. 

Thus,  (a+b)c=ac+bc. 

For  a  very  excellent  treatment  of  these  laws,  the  reader  is  . 
referred  to  Chrystal's  Algebra,  Part  I. 

Note.— The  establishment  of  these  thtee  great  laws  was  left  forthe  present  century, 
the  chief  contributors  thereto  being  De  Mor 
were 
Pierce 
from  those  of  ordinary  algebra. 

The  student  who  would  become  proficient  in  mathematics 
should  make  himself  familiar  with  ordinary  algebra,  for  it  is  the 
basis  of  all  advanced  mathematical  subjects.  For  example,  in 


416 


FINKEL'S   SOLUTION   BOOK. 


analytical  geometry,  the  subject  matter  is  geometry  while  the 
language  is  algebraic;  in  the  calculus,  the  subject  matter  may 
be  physics,  astronomy,  or  political  economy  while  the  language 
is  algebraic.  We  shall  solve  a  few  problems  in  algebra,  leaving 
the  student  to  gain  a  thorough  knowledge  of  the  subject  by  a 
study  of  such,  works  as  Chrystal's  Algebra,  2  vols. 

I.  An  estate  was  divided  among  three  persons  in  such  a 
way  that  the  share  of  the  first  was  three  times  that  of  the  sec- 
ond, and  the  share  of  the  second  twice  that  of  the  third.  The 
first  received  $900  more  than  the  third.  How  much  did  each 
receive?  [From  Hall  and  Knight's  College  Algebra,  p.  69, 
prob  40.} 

1.  I^et  ^r=the  number  of  dollars  in  the  share  of  the 

third  person.     Then 

2.  2^=the  number  of  dollars  in  the  share  of  the  sec- 

ond, and 

3.  6^r=3X2^r=the  number  of  dollars  in  the  share  of 

the  first. 

4.  Qx— .#=5.*=the  number  of  dollars  the  first  received 

more  than  the  second. 

II.  -I  5.     900= the  number  of    dollars    the    first  received 
more  than  the  second. 

6.  .'.  5^=900, 

7.  x=\  of  900=180,  the  number  of  dollars  in  the 

share  of  the  third, 

8.  2^=360,  the   number  of  dollars   in  the  share  of 

the  second, 

9.  6^=1080,  the  number  of  dollars  in  the  share  of 

the  first. 

(   $180=share  of  the  third, 
III./.  <'    $360=     "     "     "   second,  and 
($1080=     "     "     "   first. 

I.  The  length  of  a  room  exceeds  its  breadth  by  8  feet;  if 
each  had  been  increased  by  2  feet,  the  area  would  have  been  in- 
creased by  60  square  feet;  find  the  original  dimensions  of  the 
room.  [  From  Hall  and  Knight's  College  Algebra,  p.  69,  prob.  jj.] 

f  1.     L,et  ^=the  number  of  feet  in  the  breadth  of  the 
room.     Then 

2.  ^-j-8=the  number  of  feet  in  the  length,  and 

3.  (x+8)x=x'2-\-8x=the  number  of  square  feet  in  the 

area. 

4.  jr+2=the  conditional  number  of  feet   in  the 

breadth,  and 

5.  ^-j-2+8=^-|-10=the  conditional  number  of  feet  in 

the  length.     Then 


ARTHUR  CAYLEY. 


ALGEBRA. 


417 


•II.  •{  6.     (#4-2)  (#-1-10)=:  #2-|-12#-{-20:= the    conditional    num- 
ber of  square  feet  in  the  area. 

7.  (#2+12#4-20)-(#2+8#)=4#+20=the  number  of 

square  feet  in  the  increase  in  the  area, 

8.  60=the  number  of  square  feet  in  increase  in  area. 

9.  .'.   4#4-20rr:60, 

10.  4#=40,  by  subtracting  20  from  both  sides. 

11.  #=£  of  40=10,  the  number  of  feet  in  the  breadth, 

and 
^12.     #4-8=:  18,  the  number  of  feet  in  the  length. 


Ill 


.     1 10  feet=the  breadth,  and 
'  '  '    (18  feet=the  length. 


I.     A  takes  3  hours  longer  than  B  to  walk  30  miles;  but  if 
'he  doubles  his  pace,  he  takes  2  hours  less  time  than  B;    find 
their  rates  of  walking.     [From  Hall  and  Knight's  College  Alge- 
' 


II. 


1. 

L,et  #=A's  rate  in  miles  per  hour,  and 

2. 

B.s     M            «        «       ,i 

3. 

30 
Then  —  =number  of  hours  it  takes  A  to 

travel  30 

X 

miles. 

4. 

30 
—  =  number  of  hours  it  takes  B  to  travel 

30  miles. 

y 

30     30 

5. 

.'.  —  —  —  =3,  by  the  first  condition  of  the 

problem. 

6. 

2.r:=A's  conditional  rate  in  miles  per  hour. 

30    15 

7. 

Then  -^—=  —  ^number  of  hours  it  takes  A  to  travel 

LX       X 

30  miles. 

30     15 

» 

8. 

/.  —  —  —  =2,  by  the  second  condition  of 

the  prob- 

lem. 

15     15      3 

9. 

_LC/          .LtJ           ZJ          —                   /(*\ 

—  —  —  =-p-,  from  (5). 

10. 

—  =  3J,  by  adding  (8)  and  (9). 

y 

1      7 

11. 

12. 

30 
/.  j/=-=-  =4^=number  of  miles  B  travels 

per  hour. 

13. 

15     15     3 
—  _—  =-^-,  by  substituting  for_y  in  (9). 

14. 

15     7      3 

T~  2  ~  2  ' 

418 


FINKEIv'S  SOLUTION  BOOK. 


«•  -' 

17.  .'.  -r=3rrnumber  of  miles  A  travels  per  hou*. 


Ill 


.     (    3  miles=A's  rate  per  hour,  and 
'  '  '   {  4|  miles=B's  rate  per  hour. 


I.  In  a  mile  race  A  gives  B  a  start  of  44  yards  and  beats 
him  51  seconds.  In  the  second  trial  A  gives  B  a  start  of  1  min- 
ute and  15  seconds,  and  is  beaten  by  88  yards.  Find  the  rate  of 
each  in  miles  per  hour.  \Todhunter*  s  Algebra,  p.  ioj,  prob.  23; 
Wentworttts  Complete  Algebra,  p.  179,  prob.  55.] 


II. 


(i. 

2. 
3. 

4. 
5. 

6. 
7. 

8. 

9. 
10. 
11. 
12. 
13. 
14. 
15. 

16. 

lyet  .z=A's  rate  in  yards  per  second. 
j^=rB's  rate  in  yards  per  second. 
1  mile=1760  yards. 

n  the  first 

in  second 

I)  by  1760. 
5)  by  1672. 
[);  whence 

per  hour, 
in  (3)  and 

cond. 

1760-44     1716 

y         y 

trial. 
.    1716     1760 

y          x 

1760-88     1672 

trial. 
.    1760     1672 

=  t&  .  .  .  .  \4). 

39       1       51 

40j/     x     1760 
20       1       75                                           f 

l%     *58167V 

760y  ~~  33440  '  ^v  subtracting  (3)  from  (' 

44 

y=—<r  yards,  B's  rate  per  second. 

3600     44 

—  =-oo,  by  substituting  the  value  of  y 

changing  the  signs. 

88 
.  *  .  X=-^F  yards=  A's  rate  in  yards  per  se 

ALGEBRA. 


-IT        -m     .,        3600     88     ,, 

17.  .'.12  miles=  ..  _     X— A s  rate  m 


419 
per  hour. 


Ill     •     I  ^  miles:=B's  rate  per  hour. 
'  '  *   1  12  miles=A's  rate  per  hour. 


per 

I.     THE   QUADRATIC   EQUATION. 
5.     ax?+6x+c=Q,  is  the  general  quadratic  equation. 


x= — 


In  the  solution  of  exercises  involving  quadratic  equations, 
students  should  be  thoroughly  grounded  in  the  method  of  com- 
pleting the  square,  and  this  method  should  not  be  superseded  by 
the  Hindoo  Method  nor  the  Method  of  Factoring,  though  this 
latter  method  should  receive  due  consideration.  When  the 
method  is  thoroughly  impressed  upon  the  mind  of  the  student 
he  should  be  encouraged  to  solve  examples  by  merely  substitut- 
ing in  the  general  formulae  above. 

Thus,  find  the  values  of  x  satisfying  the  equation 

0.     Here,  a=2,  £=5,  and  *=— 33. 


Then 


x ,  ==. — 


5+T/25-4-2—  33  __5+17 
2-2  4 


Xo=— 


5-i/25-4-2—  33          ~ 


0 


2-2 


I.     Find  the  price  of  eggs  per  score  when  10  more  in 
cents'  worth  lowers  the  price  31£  cents  per  hundred.     [  Went 
worttts  Complete  Algebra,  p.  216,  prob.  <?.] 

1.  Let  x= price  per  score. 

2.  -|j=price  per  egg. 


420 


FINKEL'S   SOLUTION   BOOK. 


II. 


3.  5*  —  price  of  100  eggs. 

4.  62^-^--=—   —  =number  of  eggs   in   62J  cents' 


worth. 


c       1250  ,  1A    1250+10*  ,        .-    1A 

5.     -  +10=  —         —  ^number  if    10    more  be 
x  x 

added. 

C01    1250+10*       25*  •         ,  .,  1A 

6-     62^     —3—    =  50Q+4;r=price  of  each  egg,  if  10 


* 


7, 


more  be  added  to  62|  cents'  worth. 

625* 

—price  of  100  eggs. 


125+* 


.'.    5*— 


625* 


, 

9.     x 


125+* 
25*    3125 


*        .  r  r          • 

clearing  of  fractions,  transpos- 


\         4 

ing,  combining,  and  dividing  through  by  the 
coefficient  of  *2. 
1A      ,    25    .  625     50625    ,  ,    . 

10.  *     "T"jr+~5T"=  ~~gi  —  >  by  completing  the  square. 

25        225 

11.  *—  —  =±—  ^—  ,  by  extracting  the  vSquare  root. 

12.  *=31J;  or  -25. 


III.  .'.  31Jc.=price  of   the  eggs  per  score.     The  negative 
value  is  inadmissible  in  an  arithmetical  sense. 

x*+y  =11     (1)) 

\    Find  the  values  of  *  and  r.   [From  Schuy- 

•   x  +y*=  7     (2)J 
ler>s  Complete  Algebra,  p.  368,  prob.  4.] 

1.  *2—  9=2—  y    (3),  by  transposing  in  (1). 

2.  *  —  3=4—  y2  (4),  by  transposing  in  (2). 

3.  --3== 


II.  < 


4. 


6.     K2~ 


=-^,  or 


y 


*+3 
1 


(6), 
=4— 


y  transposing. 
2  1 


*+3    4  (*+3) 2          *+3    4  (*+3) 2 
completing  the  square  in  (6). 
1  1 


(7),  by 


root  of  (7). 


)>  by  extractinS  the  S(luare 


ALGEBRA. 


421 


8.     .'.  y=2,  by  transposing 


in  (8). 


^7,  by  substituting  value  of  y  in  (2). 


III.  \  -* 


x  and  j  have  three  other  values  in  addition  to  those  found. 
For  a  number  of  different  solutions  giving  the  four  values  of  x 
andjy,  see  The  American  Mathematical  Monthly. 


Find  x,  y,  and  z. 


For  a  solution  of  this  example,  see  The  Mathematical  Mag- 
azine, edited  by  Dr.  Artemas  Martin,  Washington,  D.  C. 

I.  Find  two  numbers  whose  product  is  equal  to  the  differ- 
ence of  their  squares,  and  the  sum  of  whose  squares  is  equal  to 
the  difference  of  their  cubes.  [Ray's  Higher  Algebra,  p.  230, 
prob.  p.] 

Let  ^=greater  number, 

and  j/=less  number. 

xy=x*-y*  (1). 

x*+y*=x*—y*  (2). 

Let  x=ay,  then 

ay2=a2y2— y*  (4),  by  substituting  the  value  of  x 

in  (1). 

a2—  a— 1  (5),  by  dividing  (4)  byjj/2,  and  arranging, 
whence  a=|-(l=fcV  5  ,  by  completing  the  square  of 
(5),  and  extracting  the  square  root,  and  trans- 
posing. 


II.  < 


9.        y  =  -I^  = 

2(1±V  5) 
in  (2). 

10.  x=ay=m- 


substituting   the  value  of  a 


5)  (JV  5)  = 


5). 


III. 


and 


II.     INDETERMINATE  FORMS. 


6.  The  symbol,  0,  is  defined  by  the  equation  a— a=Q.  It  is 
not  used  to  denote  nothing,  but  is  used  to  denote  the  absence 
of  quantity. 

All  operations  upon  this  symbol  are  impossible.     Thus,  0X5, 


422  FINKEL'S   SOLUTION   BOOK. 

5-K),  5°,  are  all  impossible  operations.  Standing  apart  from  the 
conditions  imposed  upon  the  quantities  from  which  0  arises  by 
certain  limitations,  the  operations  above  indicated  are  absolutely 
meaningless.  But  such  indicated  operations,  and  many  others 
of  the  same  nature,  do  very  frequently  occur  in  mathematical 
investigations,  and  when  they  do  thus  arise  they  must  be  inter- 
preted in  such  a  way  as  to  conform  to  the  fundamental  laws  of 
mathematics. 

In  conformity  to  these  laws,  OX«=<2XO=0;  a+Q=Q+a=a; 
0^=0. 

The  symbol,  GO  ,  is  used  to  represent  a  quantity  that  is  larger 
than  any  assignable  quantity,  however  large. 

What  meaning  shall  be  attached  to  the  following  indicated 
operation,  #-K)?  It  is  impossible  to  perform  this  operation. 
Suppose  we  divide  a  by  h.  This  is  possible,  provided  a  and  h 

are  real  numbers,  and  is  indicated  thus,—.     Now  what  happens 

to  the  value  of  the  fraction  — ,  if  we  conceive  h  to  diminish 

indefinitely?  We  know  that  as  the  denominator  of  a  fraction 
decreases,  the  value  of  the  fraction  increases.  Hence,  if  the 
value  of  the  denominator  becomes  very  small,  the  value  of  the 
fraction  becomes  very  large.  If  the  denominator  becomes  less 
than  any  assignable  quantity,  the  value  of  the  fraction  becomes 
larger  than  any  assignable  quantity.  All  this  is  concisely  and 

accurately  stated  as  follows:  I  -7-  I— GO  ,  or  I  —  I         =00  . 

h=Q  L  k  J  L  h  J  A=0 

Hence,  the  inaccurate  though  common  form,  a-s-O,  must  be 
interpreted  in  the  light  of  the  above  explanation  and,  therefore,, 
a-*-0=oo  ,  briefly  though  inaccurately  expressed. 

tf°=l,  for  all  finite  values  of  a\  but  is  indeterminate  if  a  is  GO  . 
Oa=0,  for  all  finite  values  of  a\  but  is  infinite  if  a  is  infinite, 
fl-s-oo  =0,  for  all  finite  values  of  a;  but  is  indeterminate  if  a  is 
infinite.  0°  is  indeterminate,  oo  —GO  ,  co  -5-00  ,  CH-0,  OXoo  ,  and 
I00  are  all  indeterminate.  But  when  these  forms  occur  in  any 
mathematical  investigation,  they  usually  have  a  determinate 
value. 

^a—^-i  0  ~l  0 

Thus, .    =—. -?-=a+x  I    .   =%a.     Here,  -TT-— 2a. 

a~X    \x=a        0  \x=a  0 


All  the  other  forms  may  be  reduced  to  the  form  -^T. 

a       a      a— a      0  .  a       OXa        0 

ThUS,  00-00=  _-_=_=_;    0X00  =0  X  ^  --  -g-  =  -g- 


ALGEBRA.  42S 

a      b       a  v  0     0X0      0  ...    0 

="x=  ;  log'  (1  )= 


Since  the  log.  (1°°)  is  indeterminate,  the  quantity  1s0  is  indeter- 
minate. 

It  is  important  that  the  student  masters  the  meaning  of  these 
forms,  as  they  occur  very  frequently  in  the  higher  mathematics. 
For  example,  the  Differential  Calculus  rests  largely  upon  the 

proper  interpretation  of  -rr. 


r  . 

1.     Find  the  limiting  value  of  —  —  ^  —  —  r-r,  when  x=2 

x'2—  HbH-16 

I  0      x-2x-3~l 


-_i 

0     (*-2)(.r-8)_=2"  ^-S^=2    "-6 

^2-1-2^ 

2.     Find  the  limiting  value  of  ~          -  when  x=Q  and  when 


0  _  _  « 

0     jf 


3.     Find  the  limiting  value  of    8_-|o    .  ^  when  jr=2  and 
when  JT=QO  . 


00 


/y/  ^J  —  y  g. 

4.     Find  the  limiting  value  of  —  =  -  ^=z  when  x—  a* 

r  «  =F  ^ 


424  FINKEIv'S  SOLUTION   BOOK. 

Let  x=a+k,  where  h  approaches  0  as  a  limit.     Then 


.-    A  2+etc. 


-%— o 

EXAMPLES. 

i 

X 


1.     Find  the  limiting  value  of       .,  when  jr=0. 


r  —  V^  —  a 

2.  Find  the  limiting  value  of  -  -  when  x=a. 

x—a 

]_  _  ^"^ 

3.  Find  the  limiting  value  of  --  g-=  when  ^=1. 

1—  v  x 


4.     Find  the  limiting  value  of    2       when  x=—~L. 
.5.     a"-^1       =what? 

=what? 


x  —  a 


.  x 
7  =what? 


8.  _  =what 


III.     ARITHMETICAL  FALLACIES. 
7.     First  Fallacy.  —  Assume  that  a=b.     Then 


.'  .  b—a-\-b 

.:  b=Zb. 
.'.  1=2, 


ALGEBRA.  425 

8.     Second  Fallacy.  —  Let  a  and  b  be  two  unequal  numbers, 
and  let  c  be  their  arithmetical  mean.     Then 


.'.  a—c—b—c 
.'.  a=b. 

Since  we  assumed  that  a  and  b  were  unequal,  where  is  the  fallacy 
in  our  reasoning? 

9.  Third  Fallacy.  —  We  have  (—  1)2=1.    Taking  logarithms, 

2  log.  (-l)=log.  1=0 
.-.log.  (-1)=0 
.-.  —  l=<?o 
.'.  —1=1,  since  ^°=1. 

10.  Fourth  Fallacy.  —  We  know  that 

log.  (l+^)=^-i^a+i^8-  ---- 

If  x—\,  the  resulting  series  is  convergent;  hence 

log.  2=l-J+J-i+i-i4^-i-H-  .... 
2  log.  2-2-l+f-^+f-J+|-i+-|-  ..... 

Taking  those  terms  together  which  have  a  common  denomina- 
tor, we  obtain 

2  log.  2=l+i-J+i+ 


-log.  2. 
.-.2=1. 

11.  Fifth  Fallacy.  —  We  can  write  V11"!  =  V— ~\  in  the 
form 

— =J— 

V^T_  vT~ 
'  v~i     iFT 

.-.  -1=1, 

12.'  5V.rM  Fallacy,  —r  The  mathematical  theory  of  probability 
leads  to  various  paradoxes;  of  these,  one  specimen  follows:  Sup- 
pose three  coins  to  be  thrown  up  and  the  fact  whether  each  comes 
down  head  or  tail  to  be  noticed.  The  probability  that  all  the 
coins  come  down  head  is  (J)*,  that  is,  £;  similarly,  the  probabil- 
ity that  all  three  coins  come  down  tail  is  ^;  hence,  the  probabil- 
ity that  all  come  down  alike,  that  is,  either  all  of  them  heads  or 
all  of  them  tails,  is  J.  But,  of  three  coins,  thus  thrown  up,  at. 


426  FINKEVS  SOLUTION  BOOK. 

least  two  must  come  down  alike.  The  probability  that  the  third 
comes  down  head  is  -J  and  the  probability  that  it  comes  down 
tail  is  ^.  Thus  the  probability  that  it  comes  down  the  same  as 
the  other  two  is  ^.  Hence,  the  probability  that  all  the  coins 
come  down  alike  is  J. 

~IV.     PROBABILITY. 

13.  Definition.     If  an  event  can  happen  in  a  ways  and 
fail  in  b  ways,  and  each  of  these  ways  is  equally  likely,  the 

probability,  or  the  chance  of  its  happening  is  -  ^-r-,  and  the 
chance  of  its  failing  is  —  rr. 

d-TO 

That  is,  the  probability  of  an  event  happening  is  the  number 
of  favorable  ways  the  event  can  happen  divided  by  the  total 
number  of  ways  it  can  happen,  and  the  probability  of  its  failing 
is  the  number  of  ways  the  event  can  fail  divided  by  the  total 
number  of  ways  it  can  happen. 

For  example,  if  in  a  lottery  there  are  6  prizes  and  23  blanks, 
the  chance  that  a  person  holding  1  ticket  will  win  a  prize  is  T6F, 
and  his  chance  of  not  winning  is  ff  . 

14.  The  reason  for  the  above  definition  may  be  made  clear 
by  the  following  considerations  : 

If  an  event  can  happen  in  a  ways  and  fail  to  happen  in  b 
ways,  and  all  these  ways  are  equally  likely  to  occur,  we  can 
assert  that  the  chance  of  its  happening  is  to  its  chance  of  failing 
as  a  to  b.  Thus  if  the  chance  of  its  happening  is  represented 
by  ka,  where  k  is  an  undetermined  constant,  then  the  chance  of 
its  failing  is  kb. 

.'.  Chance  of  happening  +  chance  of  failing  =  k(a-\-b).  Now 
the  event  is  certain  to  happen  or  to  fail;  therefore,  the  sum  of 
the  chances  of  happening  and  failing  must  represent  certainty. 
If,  therefore,  we  agree  to  take  certainty  as  our  unit,  we  have 


=,  or    = 
~ 

.*.  the  chance  that  the  event  will  happen  is  -^rr»  an(i  the 

chance  the  event  will  not  happen  is—  r-r. 

15.  The  subject  of  probability,  from  the  mathematical  point 
of  view,  is  a  very  difficult  one.  That  it  is  a  very  important  sub- 
ject, no  one  will  deny  after  having  read  Jevons's  Principles  of 
Science,  2  vols.,  in  the  first  volume  of  which  he  has  given  con- 
siderable attention  to  its  treatment.  Our  space  is  too  limited  to 
give  here  more  than  a  passing  reference  to  the  subject.  Those 
who  desire  to  study  the  subject  thoroughly,  should  read  Tod- 


ALGEBRA. 


427 


hunter's  History  of  the  Theory  of  Probability;  LaPlace's  Theorie 
Analytique  des  Probabilities,  1812  (the  most  exhaustive  treat- 
ment of  the  subject  ever  written);  and  Whitworth's  Choice  and 
Chance.  The  last  named  book  has  a  large  number  of  problems 
worked  out  in  full. 

EXAMPLES. 

1.  I.  What  is  the  chance  of  throwing  a  number  greater 
than  4  with  an  ordinary  die  whose  faces  are  numbered  from  1 
to  6? 

6=number  of  ways  the  die  can  fall. 

2r=number  of  ways  the  die  can  fall  so  as  to  give  a 

number  greater  than  4,  viz.,  5  and  6. 
3.     .*.  •§•=£= the  required  chance,  by  definition. 


II. 


III.     /.  the  required  chance  is  J. 

2.  I.     Find  the  chance  of  throwing  at  least  one  ace  in  a 
single  throw  with  two  dice. 

1.  6=number  of  ways  one  of  the  dice  may  fall. 

2.  6X6=36=:number  of  ways  the  two  dice  may  fall, 

since  with  each  of  the  six  ways  the  first  may 
fall,  there  are  six  ways  in  which  the  second 
may  fall. 

3.  5=number  of  ways  one  die  may  be  thrown  with- 

out the  ace  coming  up. 

4.  25=5X5=number  of  ways  the  two  dice  can  be 

thrown  without  either  of  them  being  ace. 

5.  .*.  ffr^the  chance  of  not  throwing  an  ace. 

6.  .'.  1—  £f=iJ=the  chance  of  throwing  at  least  one 

ace. 

III.     .*.  the  chance  of  throwing  at  least  one  ace  is  ££. 

3.  I.     If  four  coins  are  tossed  find  the  chance  that  there 
should  be  two  heads  and  two  tails. 


II. 


J^chance  of  head  or  tail  with  one  coin. 
^==(jj*=chanice  of  all  heads,  tossing  4  coins. 
-j^=(^)4=chance  of  all  tails,  tossing  4  coins. 
i~  f^i(i)3—  chance  of  1  head  and  3  tails. 
J=iX(J)*(J)=chance  of  1  tail  and  3  heads. 


r  1. 
2. 
3. 
4. 
5. 


. 
6-     t^?i)  2  (i)  2  -chance  of  2  heads  and  2  tails. 


II 


III.     .'.  the  chance  of  throwing  2  heads  and  2  tails  is  f. 

4.     I.     A  bag  contains  5  white,  7  black,  and  4  red  balls;  find 
the  chance  that  three  balls  drawn  at  random  are  all  white. 


428 


FINKEIv'S  SOLUTION  BOOK. 


.     16X15X14=number  of  ways  3  objects  can  be  se- 
lected from  16  objects. 

.     5X4X3=number  of  ways  3  objects  can  be  selected 
,  from  5  objects,  which  is  the  number  of  ways  the 

3  white  balls  may  be  selected  from  the  5  white 
balls. 
5X4X3        (       . 

'     •'•  16X15X14=A=the 
III.     .'.  the  required  chance  is 


PROBLEMS. 

1.  State  as  a  theorem  the  fact  implied  in  the  following  equa- 
tions:     92  —  72  =  4-8;     52  -  -  32  =  4.4;     932  -  -  9i2  =  4-92; 
32  —  j2  _  4.2      Prove  it,  and  then  express  the  theorem  in  its 
most  general  terms. 

2.  How  find  the  highest  common  factor  of  two  polynomials 
that  cannot  be  readily  factored?     Prove  your  answer.     Illustrate 
by  finding  the  H.  C.  F.  of 

6x*  +  7^2  —  $x  and  15^*  +  3I*3  +  io^r2. 

3.  A  cistern  can  be  filled  in  4  hours  by  two  pipes  running 
together,  and  in  6|  hours  by  one  of  the  pipes  alone.     In  how 
many  hours  can  the  other  pipe  fill  the  same  cistern? 

4.  If  V  a  and  V  b  are  surds,  prove  that  V  a  ±  V  b  cannot  be 
a  rational  number. 


5 .     Simplify 


+  y 


2         TT — 


—  1/2 


Check  your  work  by  substituting  x  —  4  and  y  =  I,  both  in  the 
given   expression   and   in   the   simplified   form,   and   comparing 
results. 
6.     Given  the  two  simultaneous  equations 

find  all  the  pairs  of  values  of  x  and  y  that  satisfy  them. 

Cornell  University  —  Entrance  Examination,  1899. 

T. 


i.     Resolve 


—  3*  — 


into  partial  fractions. 


2.     At  an  election  there  are  4  candidates,  and  3  members  to 


PROBLEMS,  429 

be  elected,  and  an  elector  may  vote  for  any  number  of  candidates 
not  greater  than  the  number  to  be  elected.  In  how  many  ways 
may  an  elector  vote  ? 

2 

3.  Find,  by  using  logarithms,  the  value  of   V41.72  X  (.054)3. 

4.  Show  that  for  any  two  quantities,  the  square  of  their  geo- 
metric mean  is  equal  to  the  product  of  their  arithmetic  and  har- 
monic means. 

5.  Draw  the  graph  of  the  function  xz  +  ,r2  +  x — 100;    and 
find  the  root  between  4  and  5,  correct  to  three  places  of  decimals, 
of  the  equation  ,r3  -}-  .r2  -f-  x  —  100  =  o. 

6.  If  h,  k,  are  the  roots  of  ax*  +  bx  +  c  —  o,  find  the  value 
,   1       1 

°f^+^" 

7.  The  square  of  x  varies  as  the  cube  of  y,  and  x  =  3  when 

1 

y  —  4.     Find  the  value  of  y  when  x  —  ~~rw- 

V  o 

Cornell  University  —  Entrance  Examination,  1899. 

I.     (a)   Simplify  the  expression 

Z6e  _  26c     _ 

J+~^    b    •  J+7 
i      H~  i         i 


1 


Show  how  the  symmetry  of  this  expression  may  be  made  to  serve 
as  a  partial  check  upon  the  result. 

(b)  As  x  becomes  more  and  more  nearly  equal  to  2,  what 

value   does   the    fraction   approach?     What   is   the 

value  of  this  fraction  when  x  —  2? 

2.  The  sum  of  the  three  digits  of  a  number  is  9;   the  digit 
in  hundreds'  place  equals  J  of  the  number  composed  of  the  two 
other  digits,  and  the  digit  in  units'  place  equals  -J  of  the  number 
composed  of  the  two  other  digits ;  what  is  the  number  ? 

3.  Prove  that :    (o)  x*  +  yn  is  exactly  divisible  by  x  +  y  if  n 
is  any  odd,  positive  integer  whatever. 

(b)   If  a  +  b  is  constant,  then  a-b  is  greatest  when  a  =  6. 

4.  (a)   State  what  seem  to  you  to  be  the  chief  differences 
between  algebra  and  arithmetic. 

(b)  Define  negative  number,  subtraction,  and  multiplica- 
tion, and  show  how,  from  your  definition,  the  following  rules  may 
be  deduced:  (i)  "change  the  sign  of  the  subtrahend  and  pro- 
ceed as  in  addition";  (2)  give  the  product  the  positive  or  the 


430  FINKEL'S    SOLUTION    BOOK. 

negative  sign  according  as  the  two  factors  have  like  or  unlike 
signs." 

5.  Given  the  equation  ax2  +  bxy  +  cyz  =  o  in  which  a,  b, 
and  c  are  real,  —  the  ratio  x\y  being  unknown.  Find  the  sum 
and  also  the  product  of  the  roots  (i.  e.  of  the  values  of  the  ratio 
x:y)  of  this  equation.  If  a  approaches  o  relatively  to  b  and  c, 
what  happens  to  these  roots?  For  what  relative  values  of  a,  b, 
and  c  are  the  two  roots  equal?  One  twice  the  other?  Both 
imaginary  ?  T. 

Cornell  University  —  Scholarship  Examination,  1899. 

1.  Two  men,  A  and  B,  had  a  money-box  containing  $210, 
from  which  each  drew  a  certain  sum  daily ;  this  sum  being  fixed 
for  each,  but  different  for  the  two.     After  six  weeks,  the  box 
was  empty.     Find  the  sum  which  each  man  drew  daily  from  the 
box ;    knowing  that  A  alone  would  have  emptied  it  five  weeks 
earlier  than  B  alone. 

Obtain  two  solutions,  and  interpret  the  negative  answer. 

2.  Solve  the  equation 

x  +  b       2a    _., 2a  /-,    2a  ~  6\ 

$a  -X—.-6    '    ~~~6~\      x  —  b) 

3.  Reduce  to  their  lowest  common  denominator 

i                                              i 
and  


-  6  4-3T3  -  6x2  -  4^  +  6 

and  find,  and  reduce  to  their  lowest  terms,  the  difference  and  the 
quotient  of  these  two  fractions. 
4.     Write  out  (x  —  y)9. 

(4  Vbz  —  1\  9 

o   g  --  \aVa.b  T  }  ;  reducing  the  an- 

swer to  the  simplest  possible  form,  in  which  it  is  free  from  nega- 
tive and  fractional  exponents,  and  has  only  one  radical  sign. 
Harvard  University  —  Entrance  Examination, 

1.  Resolve  into  three  factors  (x2  —  x)B  —  8. 

2.  Find  the  greatest  common  factor  of 

x*  +  5*z  +  6>  and  3^.+  7^  +  3^  +  2. 


Solve  the  equation  V^r-4  +  V^—  11-  V2*9=0. 


Vr 
*J  x  — 


4.     Simplyfy 

x-y 


PROBIvEMS.  431 

5.  Solve  the  simultaneous  equations 

3O2  +  xy)  —  403;,  x  —  y  —  2. 

6.  What  is  the  geometrical  mean  between 

2x  —  3  and  zx*  +  x*  —  ^x  —  3  ? 

7.  A  and  B  start  at  the  same  time  from  the  same  point  in  the 
same  direction.     A  goes  at  the  uniform  rate  of  60  miles  per  day  ; 
B  goes  14  miles  the  first  day,  16  miles  the  second  day,  18  miles 
the  third  day  and  so  on.     At  the  end  of  50  days  who  will  be  ahead, 
and  by  how  much? 

Massachusetts  Institute  of  Technology  —  Entrance  Ex- 
amination in  Advanced  Algebra,  1892. 

1.  (a)   Show  that 

(w+i)    (n  +  2)    (n  +  3)  n 

-_i  =—(n*  +  6  n+  n). 
2  +  3  6 

(b)  Find  the  algebraic  expression  which  when  divided 
by  x2  —  2X  -{-  i  gives  a  quotient  x2  -\-  2x  +  i  and  a  remainder 
x  —  i. 

2.  (a)   Reduce   to   a   common   denominator    (arranging  the 
terms  of  the  numerator  according  to  ascending  powers  of  x) 

ABC  D 

-  +  -  +  --  h  - 
x      x+i       O+l)2       (*+i)3 

(  b  )  Having  given  that  :  A  =  2,  B  —  2  A  —  o,  C  —  2B 
-+3A=  —  3,  D  —  2C  +  38  =_i,£_  2D  +  3c=0;  find  the 
values  of  B,  C,  D,  and  E. 

3.  Solve  for  x  and  y,  2X  —  3^,  +  14  =  o,  53;  —  ^x  =  26. 

4.  (a)   Simplify  j&+9xt-4r(x*)9-*r+x&-*r. 

(*)  Multiply    xn+Jt+lbyx-a+J+l. 

5.  Solve  for  x    (a)    x^  -\-  2a?  =  $ax.     Also 

(^)/  +  3^  =  4. 

6.  Solve  the  simultaneous  equations 


Also  (&)    (x  +  y)  =  20,  x*  +  y2  =  2a2. 

Princeton  University  —  Entrance  Examination,  1893. 

I.     Write  the  factors  of  the  following  expressions: 
x^  —  (x  —  6)2  and  w6  —  64W6. 


432  FINKEL'S    SOLUTION    BOOK. 

12 


2.     Simplify  x  —  i 


12 

—  5 


3.  A  and  B  can  do  a  piece  of  work  in  m  days  ;   B  and  C  can 
do  it  in  n  days  ;  C  and  D  in  p  days  ;  and  D  and  A  in  r  days.     In 
how  many  days  can  all  working  together  do  it  ? 

4.  Multiply  x  +  y  V—  z  by  y  —  2  ^—x. 

5.  Solve  the  equation  y2  +  2  (a  -f  6)  y  =  —  i8a<. 

6.  Extract  the  square  root  of 


1 


7.  Simplify. 

l+T 

8.  Write  the  6th  term  of  (a  —  2&)J. 

Yale  University  —  Entrance  Examination,  1893. 

1.  (a)   Solve  the  equation  ax2  -j-  for  -f-  c  =  o. 

(fr)  What  relation  must  exist  between  its  coefficients  in 
order  (i)  that  its  roots  may  be  imaginary,  (2)  that  they  may 
be  real  and  equal,  (3)  that  they  may  be  real  and  unequal? 

(c)  If  the  coefficient  a  diminishes  without  limit,  what 
limits,  if  any,  do  the  roots  respectively  approach  ? 

2.  Make  the  first  members  of  the  following  equations  perfect 
squares,  without  introducing  fractions: 

2^2  _  ^x  —  2j  3^2  _  g^  —  _  4 

3.  Solve  completely  the  simultaneous  equations  x2  -j-  xy  +  y2 
:=  19,  x2  —  xy  +  3>2  =  7,  and  group  distinctly  the  corresponding 
values  of  x  and  y. 

4.  Convert  3.14159  into  a  continued  fraction,  and  obtain  four 
convergents.     What  is  the  limit  of  the  error  in  taking  the  fourth 
convergent  for  the  value  of  the  decimal  ? 

5.  (a>)  Derive  the  formula  for  the  number  of  permutations 
of  m  things  taken  n  at  a  time. 

(b)  From  10  different  things  in  how  many  ways  can  a 
selection  of  4  be  made  ? 


PROBLEMS.  433 

6.     (a)    Write    equivalents    for    the    following    expressions: 
log  «1;   log  ««;   log  flO,  if  a>l ;   log  a2  -  log  £2 ; 


(b)  Given  the  mantissa  of  log    10257  =  0.40993,  to  find 

log10T/0257: 

7.     Given  log  aN  and  log  ab,  to  find  log  bN. 

Sheffield  Scientific  School  of  Yale  University— Entrance 
Examination  in  Advanced  Algebra,  1892. 


434 


FINKEL'S    SOLUTION    BOOK. 


PROBABILITY  PROBLEMS. 


I.  If  three  pennies  be  piled  up  at  random  on  a  horizontal 
plane,  what  is  the  probability  that  the  pile  will  not  fall  down? 

The  pile  will  stand  if  the  common  center  of  gravity  of  the  sec- 
ond and  third  coins  falls  on  the  surface  of  the  first  or  bottom, 
coin. 

Let  r  be  the  radius  of  a  penny  ;  then  the  center  of  the  second 
coin  may  fall  anywhere  in  a  cir- 
cle whose  radius  is  2r  and  cen- 
ter the  center  of  the  surface  of 
the  first  or  bottom  coin,  and  the 
center  of  the  third  coin  may  fall 
anywhere  in  a  circle  whose  ra- 
dius is  2r  and  the  center  the 
center  of  the  surface  of  the  sec- 
ond coin.  The  number  of  posi- 
tions of  the  center  of  the  second 
coin  is  therefore  proportional  to 
47Tr2,  and  for  every  one  of  these 
positions  the  center  of  the  third 
coin  can  have  4?rr2;  hence  the 
total  number  of  positions  of  the  ^G'  s' 

second  and  third  coins  is  proportional  to  167rV4. 

We  must  now  determine  in  how  many  of  these  167rV4  posi- 
tions the  pile  will  stand. 

Let  A  be  the  center  of  the  first  or  bottom  coin,  and  B  the  cen- 
ter of  the  second  coin.  Take  AD=AB,  and  with  center  D 
and  radius  2r  describe  the  arc  CHJ.  If  the  center  of  the  third 
coin  is  on  the  surface  CFJH,  the  second  and  third  coins  will  re- 
main on  the  first,  since  BN=NH,  BT=TC,  and  the  pile  will 
not  fall  down. 

When  AB  is  not  greater  than  -Jr,  the  circle  CHJ  will  not  cut 
the  surface  of  the  second  coin,  and  the  pile  will  stand  if  the  cen- 
ter of  the  third  coin  is  anywhere  on  the  second. 


Let  AJ3=AD=x,S=surface  CFJH,   and  /=the    probability 
required  ;    then  DB=%x,  BG=Jtr*~^X' 


,  arc  Cf=r  cos"1 


PROBABILITY  PROBLEMS.  435 


,  and 


1  1 


/.r  /•  /''Sr2  _  4*2>\ 

-  /    (  S—  nr*  )xdx.  lr*  cos-1  (  *r  A  }xdx, 

V/  J  ^      ±rx      y 


(K^2  _  4v2x      1 
-14?F->)-^(5r2- 


NOTE.  —  This  solution  is  due  to  Artemas  Martin,  M.  A.,  Ph.  D.,  LL.  D., 
member  of  the  London  Mathematical  Society,  member  of  the  Edinburg 
Mathematical  Society,  member  of  the  Mathematical  Society  of  France, 
member  of  the  New  York  Mathematical  Society,  member  of  the  Philo- 
sophical Society  of  Washington  and  Fellow  of  the  American  Association 
for  the  Advancement  of  Science,  Washington,  D.  C.,  who  is  one  of  the 
great  peers  of  mathematical  science. 


436  FINKEL'S    SOLUTION    BOOK. 


BIOGRAPHY. 

ARTEMAS  MARTIN,  M-  A.,  PH.  D.,  LL.  D. 


This  eminent  mathematician  was  born  in  Steuben  county,  N.  Y.,  Au- 
gust 3,  1835-  Early,  his  parents  moved  to  Venango  county,  Pa.,  where  they 
lived  for  many  years.  Dr.  Martin  had  no  schooling  in  his  early  boyhood, 
except  a  little  primary  instruction;  but  by  self-application  and  indefatiga- 
ble energy  which  have  told  the  story  of  many  a  great  man,  he  has  become 
familiar  to  every  mathematician  and  lover  of  science  in  every  civilized 
country  of  the  world 

He  was  never  a  pupil  at  school,  except  when  quite  small,  until  in  his 
fourteenth  year.  He  had  learned  to  read  and  write  at  home,  but  knew 
nothing  of  Arithmetic.  At  fourteen  he  commenced  the  study  of  Arithme- 
tic, and  after  spending  two  winters  in  the  district  school,  he  commenced 
the  study  of  Algebra.  At  seventeen,  he  studied  Algebra,  Geometry,  Nat- 
ural Philosophy,  and  Chemistry  in  the  Franklin  Select  School,  walking 
two  and  one-half  miles  night  and  morning.  Three  years  after,  he  spent 
two  and  one-half  months  in  the  Franklin  Academy,  studying  Algebra  and 
Trigonometry.  This  finished  his  schooling.  He  taught  district  schools 
four  winters,  but  not  in  succession.  He  was  raised  on  a  farm,  and  worked 
at  farming  and  gardening  in  the  summer;  chopped  wood  in  the  winter; 
and  after  the  discovery  of  oil  in  Venango  county,  worked  at  drilling  oil 
wells  a  part  of  his  time,  always  devoting  his  "spare  moments"  to  study. 

In  the  spring  of  1869,  the  family  moved  to  Erie  county,  Pa.,  where  he  re- 
sided until  he  entered  the  U.  S.  Coast  Survey  Office  in  1885.  While  in 
Erie  county,  after  1871,  he  was  engaged  in  market-gardening,  which  he 
carried  on  with  great  care  and  skill.  He  began  his  mathematical  career 
when  in  his  eighteenth  year,  by  contributing  solutions  to  the  Pittsburg 
Almanac,  soon  after  contributing  problems  to  the  "Riddler  Column"  of  the 
Philadelphia  Saturday  Evening  Post,  and  was  one  of  the  leading  contribu- 
tors for  twenty  years. 

In  the  summer  of  1864  he  commenced  contributing  problems  and  solu- 
tions to  Claris  School  Visitor,  afterward  the  Schoolday  Magazine,  pub- 
lished in  Philadelphia.  In  June,  1870,  he  took  charge  of  the  "Stairway  De 
partment"  as  editor,  the  mathematical  department  of  which  he  had  con- 
ducted for  some  years  before.  He  continued  in  charge  as  mathematial  edi- 
tor till  the  magazine  was  sold  to  Scribner  &  Co.,  in  the  spring  of  1875,  at 
which  time  it  was  merged  into  ''6Y.  Nicholas" 

In  September,  1875,  he  was  chosen  editor  of  a  department  of  higher 
mathematics  in  the  Normal  Monthly,  published  by  Prof.  Edward  Brooks, 
Millersville,  Pa.,  and  held  that  position  till  the  Monthly  was  discontinued 
in  August,  1876.  He  published  in  the  Normal  Monthly  a  series  of  sixteen 
articles  on  the  Diophantine  Analysis. 

In  June,  1877,  Yale  College  conferred  on  him  the  honorary  degree  of 
Master  of  Arts  (M.  A.)  In  April,  1878,  he  was  elected  member  of  the  Lon- 
don Mathematical  Society.  In  June,  1882,  Rutgers  College  conferred  on 
him  the  honorary  degree  of  Doctor  of  Philosophy  (Ph.  D.)  March  7, 1884, 
he  was  elected  a  member  of  the  Mathematical  Society  of  France.  In  April, 
1885,  he  was  elected  a  member  of  the  Edinburgh  Mathematical  Society. 
June  10,  1885,  Hillsdale  College  conferred  on  him  the  honorary  degree  of 
Doctor  of  Laws  (LL.  D.)  February  27,  1886,  he  was  elected  a  member  of 
the  Philosophical  Society  of  Washington.  In  June,  1881,  he  was  elected 
Professor  of  Mathematics  of  the  Normal  School  at  Warrensburg,  Mo.,  but 
did  not  accept  the  position.  November  14,  1885,  Dr.  Martin  was  appointed 


PROBABILITY  PROBLEMS. 


437 


Librarian  in  the  office  of  the  U.  S.  Coast  and  Geodetic  Survey.  On  August 
'26,  1890,  he  was  elected  a  Fellow  of  the  American  Association  for  the  Ad- 
vancement of  Science.  On  April  3,  1891  he  was  elected  a  member  of  the 
New  York  Mathematical  Society. 

All  these  honors  have  been  worthily  bestowed  and  the  Colleges  and  So- 
cieties conferring  them  have  done  honor  to  themselves  in  recognizing  the 
merits  of  one  who  has  become  such  a  power  in  the  scientific  world  through 
his  own  efforts. 

He  has  contributed  fine  problems  and  solutions  to  the  following  journals 
of  the  United  States:  School  Visitor,  Analyst,  Annals  of  Mathematics, 
Mathematical  Monthly,  Illinois  Teacher,  I oiva  Instructor,  National  Edu- 
cator, Tates  County  Chronicle,  Barnes'  Educational  Monthly,  Wittenberger, 
Maine  Farmers'  Almanac,  Mathematical  Messenger,  and  Educational 
Notes  and  Queries,  American  Mathematical  Monthly.  Besides  other  con- 
tributions, he  contributed  thirteen  articles  011  "Average  "  to  the  Mathemat- 
ical Department  of  the  Wittenberger,  edited  by  Prof.  William  Hoover. 
These  are  believed  to  be  the  first  articles  published  on  that  subject  in 
America. 

Dr.  Martin  has  also  contributed  to  the  following  English  mathematical 
periodicals :  Lady's,  and  Gentleman's  Diary,  Messenger  of  Mathematics, 
and  The  Educational  Times  and  Reprint. 

The  Reprint  contains  a  large  number  of  his  solutions  of  difficult  "Aver- 
age "  and  "Probability"  problems,  which  are  master-pieces  of  mathematical 
thought  and  skill,  and  they  will  be  lasting  monuments  to  his  memory.  His 
style  is  direct,  clear  and  elegant.  His  solutions  are  neat,  accurate  and  sim- 
ple. He  has  that  rare  faculty  of  presenting  his  solution  in  the  simplest 
mathematical  language,  so  that  those  who  have  mastered  the  elements  of 
the  various  branches  of  mathematics  are  able  to  understand  his  reasoning. 

Dr.  Martin  is  now  ( 1899 )  editor  of  the  Mathematical  Magazine,  and 
The  Mathematical  Visitor,  two  of  the  best  mathematical  periodicals  pub- 
lished in  America.  These  are  handsomely  arranged  and  profusely  illustra- 
ted with  very  beautiful  diagrams  to  the  solutions,  he  doing  the  typesetting 
with  his  own  hand.  The  typographical  work  of  these  journals  is  said  to  be 
the  finest  in  America.  The  best  mathematicians  from  all  over  the  world 
contribute  to  these  two  journals.  The  Mathematical  Visitor  is  devoted  to 
Higher  Mathematics,  while  The  Mathematical  Magazine  is  devoted  to  the 
solutions  of  problems  of  a  more  elementary  nature.  All  solutions  sent  to 
Dr.  Martin  receive  due  credit,  and  if  it  is  possible  to  find  room  for  them  the 
solutions  are  all  published.  He  has  thus  encouraged  many  young  aspirants 
to  higher  fields  of  mental  activity.  He  is  always  readv  to  aid  any  one  who 
is  laboring  to  bring  success  with  his  work.  He  is  of  a  kind  and  noble  dis- 
position and  his  generous  nature  is  in  full  sympathy  with  every  diligent 
student  who  is  rising  to  planes  of  honor  and  distinction  by  self  application 
and  against  adverse  circumstances. 

Dr.  Martin  has  a  large  and  valuable  mathematical  library  containing 
many  rare  and  interesting  works.  His  collection  of  Amerioan  arithmetics 
and  algebras  is  one  of  the  largest  private  collections  of  the,  kind  in  this 
•country. 


438  FINKEL'S    SOLUTION    BOOK. 


I.  Find  the  average  or  mean  distance 
from  one  corner. 

Taking  the  corner  from  which  the  mean  distance  is  to  be 
found  for  the  origin  of  orthogonal  co-ordinates,  and  one  of  the- 
sides  of  the  square  for  the  axis  of  abscissa,  we  have  for  the  ele- 
ment of  the  surface  dx  dy,  and  since  this  element  is  at  a  distance: 

1   Ca  f*a 
from  the  origin,  the  average—  /      /    dxdy*J(x2-\-y%) 


.:  Average=ia[V2+log«(l+V2)]. 

NOTE.  —  This  solution  is  by  Prof.  J.  W.  F.  Sheffer,  Hagerstown,  Md.,, 
whose  name  may  be  found  attached  to  the  solutions  of  many  difficult  prob- 
lems proposed  in  the  leading  mathematical  journals  of  the  United  States. 
The  above  solution  is  taken  from  the  Mathematical  Messenger,  published 
by  G.  H.  Harville,  Simsboro,  La. 

I.  All  that  is  known  concerning  the  veracities  of  two  witnesses, 
A  and  B,  is  that  B's  statements  are  twice  as  reliable  as  A's.. 
What  is  the  probability  of  the  truth  of  the  concurrent  testimony 
of  these  two  witnesses? 

Let  #=the  probability  of  the  truth  of  any  one  of  A's  state- 
ments ;  then  2#=the  probability  of  any  one  of  B's  statements.. 
The  event  did  occur  if  both  witnesses  tell  the  truth,  the  proba- 
bility ot  which  is  xX^x=2x2.  The  event  did  not  occur  if  both 
testify  falsely,  the  probability  of  which  is  (1  —  #)x(l  —  2ff)=l 
—  3#-|-2tf2.  Hence,  the  probability  of  the  occurrence  of  the  event 
on  the  supposition  that  x  is  known  is 


x)  (I—  -2*)' 

Now,  as  the  veracity  of  B  can  not  exceed  unity,  the  greatest 
value  of*  is  found  by  putting  2#=1,  which  gives  #=-J;  hence, 
x  can  have  any  value  from  0  to  -J-. 

Therefore,  the  probability  in  the  problem  is 

.,  ,        * 

P  dx 


C 

Jo 


—  3)2+7* 

Let  8«—  3=y.    Then  x=$(y—  3),  dx=\dy\   the   limits  of>  are 
1  and  —  3,  and 


PROBABILITY  PROBLEMS. 


439< 


NOTE.  —  This  solution  is  taken  from  the  Mathematical  Magazine,  Vol. 
II,  p.  122.  The  solution  there  given  is  credited  to  the  author,  Prof.  William 
Hoover,  and  Prof.  P.  II.  Philbrick. 

I.  A  cube  is  thrown  into  the  air  and  a  random  shot  fired 
through  it;  find  the  chance  that  shot  passed  through  oppo- 
site faces. 

Let  AH\)Q  the  cube.  Through  P  ',  a  point  in  the  face  EFGH  , 
draw  MK  parallel  to  HE,  and  draw  PN  perpendicular  to  HE* 
Now  if  PA  represents  the  direction  of 
the  shot,  it  will  pass  through  the  face 
ABCD,  if  it  strikes  the  face  EFGH 
anwhere  within  HMPN. 


FIG.  9. 


and  area  HMPN=u.  Then 
PK=sec  6  tan0,  7^T=tan  0 
sec0sec0,  PM=\  —  sec0tan0, 
—  tan#,  ?te(  1  —  sec#tan0)  (  1  —  tan#  )  ,the 
area  of  the  projection  of  HMPN  on  a 
plane  perpendicular  to  PA=ucos6cos(f>, 
and  that  of  EFGH=co$8cos(f). 

Since  we  are  to  consider  all  possible 
directions  of  the  shot  with  respect  to  the  cube,  the  points  of  in- 
tersection of  PA  with  the  surface  of  a  sphere  whose  center  is  A, 
and  radius  unity,  must  be  uniformly  distributed.  An  element 
of  the  surface  of  this  sphere  is  co&cfid  6  d<{).  By  reason  of  the 
symmetry  of  the  cube,  the  required  chance  is  obtained  by  finding 
the  number  of  ways  the  shot  can  pass  through  the  opposite  faces 
EFGH  and  ABCD  between  the  limits  6=0,  and  0=^  and 
0=0  and  0=tan~1  (cos#)=0',  and  the  number  of  ways  it  can 
pass  through  the  face  .S/^G^/between  the  limits  0=0  and  #=-j-;r,. 
and  0=0  and  <fc=^rt  ;  and  then  dividing  the  former  by  the  latter. 
Hence,  the  chance  required  is 


p 

r 

Jo 


=f   /?*  f 

nj°  Jo 


—  6*2  tan"1 


tan  0) 


NOTE.  —  This  solution  is  due  to  Professor  Enoch  Beery  Seitz,  member 
of  the  London  Mathematical  Society,  and  late  Professor  of  Mathematics 
in  the  North  Missouri  State  Normal  School,  Kirksville,  Mo. 


440  FINKEL'S   SOLUTION   BOOK. 

BIOGRAPHY. 


PROF.   E.  B.  SEITZ,  M.  L.  M.  S. 

Professor  Seitz,  a  distinguished  mathematician  of  his  day,  was 
born  in  Fairfield  Co.,  p.,  Aug.  24,  1846,  and  died  at  Kirksville,  Mo.,  Oct.  8, 
1883.  His  father,  Daniel  Seitz,  was  born  in  Rockingham  Co.,  Va.,  Dec.  17, 
1791,  and  was  twice  married.  His  first  wife's  name  was  Elizabeth  Hite,  of 
Fairfield  Co.,  O.,  by  whom  he  had  eleven  children.  His  second  wife's  name 
was  Catharine  Beery,  born  in  the  same  county,  Apr.  11,  1808,  whom  he 
marr-ed  Apr.  15,  1832.  This  woman  was  blessed  by  four  sons  and  three 
daughters.  Mr.  Seitz  followed  the  occupation  of  a  farmer  and  was  an  in- 
dustrious and  substantial  citizen.  He  died  near  Lancaster,  O.,  Oct.  14, 
1864,  in  his  seventy-third,  year ;  having  been  a  resident  of  Fairfield  Co.  for 
sixty-three  years. 

Professor  Seitz,  the  third  son  by  his  father's  second  marriage,  passed 
his  boyhood  on  a  farm,  and  like  most  men  who  have  become  noted,  had 
only  the  advantages  of  a  common  school  education.  Possessing  a  great 
thirst  for  learning,  he  applied  himself  diligently  to  his  books  in  private  and 
became  a  very  fine  scholar  in  the  English  branches,  especially  excelling  in 
arithmetic.  It  was  told  the  author,  by  his  nephew,  Mr.  Huddle,  that  when 
Professor  Seitz  was  in  the  field  with  a  team,  he  would  solve  problems  while 
the  horses  rested.  Often  he  would  go  to  the  house  and  get  in  the  garret 
where  he  had  a  few  algebras  upon  which  he  would  satiate  his  intellectual 
appetite.  This  was  very  annoying  to  his  father  who  did  not  see  the  future 
greatness  of  his  son,  and  many  and  severe  were  the  floggings  he  received 
for  going  to  his  favorite  retreat  to  gain  a  victory  over  some  difficult  prob- 
lem upon  which  he  had  been  studying  while  following  the  plow.  Though 
the  way  seemed  obstructed,  he  completed  algebra  at  the  age  of  fifteen,  with- 
out an  instructor.  He  chose  teaching  as  his  profession  which  he  followed 
with  gratifying  success  until  his  death.  He  took  a  mathematical  course  in 
the  Ohio  Wesleyan  University  in  1870.  In  1872,  he  was  elected  one  of  the 
teachers  in  the  Greenville  High  School,  which  position  he  held  till  1879. 
On  the  24th  of  June,  1875,  he  married  Miss  Anna  E.  Kerlin,  one  of  Darke 
county's  most  refined  ladies.  In  1879,  he  was  elected  to  the  chair  of  mathe- 
matics in  the  Missouri  State  Normal  School,  Kirksville,  Mo.,  which  posi- 
tion he  held  till  death  called  him  from  the  confines  of  earth,  ere  his  star  of 
fame  had  reached  the  zenith  of  its  glory.  He  was  stricken  by  that  "demon 
of  death,"  typhoid  fever,  and  passed  the  mysterious  shades,  to  be  numbered 
with  the  silent  majority,  on  the  8th  of  October,  1883.  On  the  llth  of 
March,  1880,  he  was  elected  a  member  of  the  London  Mathematical  So- 
ciety, being  the  fifth  American  so  honored. 

He  began  his  mathematical  career  in  1872,  by  contributing  solutions  to 
the  problems  proposed  in  the  ''Stairway"  department  of  the  Schoolday 
Magazine,  conducted  by  Artemas  Martin.  His  masterly  and  original  solu- 
tions of  difficult  Average  and  Probability  problems,  soon  attracted  universal 
attention  among  mathematicians.  Dr.  Martin  being  desirous  to  know  what 
works  he  had  treating  on  that  difficult  subject,  was  greatly  surprised  to 
learn  that  he  had  no  works  upon  the  subject,  but  had  learned  what  he  knew 
about  that  difficult  department  of  mathematical  science  by  studying  the 
problems  and  solutions  in  the  Schoolday  Magazine.  Afterwards,  he  con- 
tributed to  the  Analyst,  the  Mathematical  Visitor,  the  Mathematical  Maga- 
zine, the  School  Visitor,  and  the  Educational  Times,  of  London,  Eng. 

In  all  of  these  journals.  Professor  Seitz  was  second  to  none,  as  his 
logical  and  classical  solutions  of  Average  and  Probability  problems,  rising 
as  so  many  monuments  to  his  untiring  patience  and  indomitable  energy  and 
perseverance  will  attest.  His  name  first  appeared  as  a  contributor  to  the 
Educational  Times  in  Vol.  XVIII.,  of  Reprint,  1873.  From  that  time  until 
his  death  the  Reprint  is  adorned  with  some  of  the  finest  product  of  his 
mighty  intellect. 


BIOGRAPHY.  441 

On  page  21,  Vol.  II.,  he  has  given  the  above  solution.  This  problem 
had  been  proposed  in  1864  by  the  great  English  mathematician,  Prof.  Wool- 
house,  who  solved  it  with  great  labor.  It  was  said  by  an  eminent  mathe- 
matician of  that  day  that  the  task  of  writing  out  a  copy  of  that  solution 
was  worth  more  than  the  book  in  which  it  was  published. 

No  other  mathematician  seemed  to  have  the  courage  to  investigate  the 
problem  after  Prof.  Woolhouse  gave  his  solution  to  the  world,  till  Profes- 
sor Seitz  took  it  up  and  demontsrated  it  so  elegantly  in  half  a  page  of  or- 
dinary type,  that  he  fairly  astonished  the  mathematicians  of  both  Europe 
and  America.  Prof.  Woolhouse  was  the  best  English  authority  on  Proba- 
bilities even  before  Professor  Seitz  was  born. 

It  was  the  solution  of  this  problem  that  won  for  Professor  Seitz  the 
acknowledgment  of  his  superior  ability  to  solve  difficult  Probability  prob- 
lems over  any  other  living  man  in  the  world. 

In  studying  his  solutions,  one  is  struck  with  the  simplicity  to  which  he 
has  reduced  the  solutions  of  some  of  the  most  intricate  problems.  When 
he  had  grasped  a  problem  in  its  entirety,  he  had  mastered  all  problems  of 
that  class.  He  would  so  vary  the  condition  in  thinking  of  one  special 
problem  and  in  effecting  a  solution  that  he  had  generalized  all  similar  cases, 
so  exhaustive  was  his  analyses.  Behind  the  words  he  saw  all  the  ideas 
represented.  These  he  translated  into  symbols,  and  then  he  handled  the 
symbols  with  a  facility  that  has  rarely  been  surpassed. 

What  he  might  have  accomplished  in  his  maturer  years,  had  he  turned 
his  splendid  powers  to  investigations  in  higher  and  more  fruitful  fields  of 
mathematics,  no  man  may  say.  The  solving  of  problems  alone  is  not  a 
high  form  of  mathematical  research.  While  problem  solving  is  very  bene- 
ficial and  essential  at  first,  yet,  if  one  confines  himself  to  that  sort  of 
work  exclusively,  it  becomes  a  waste  of  time. 

He  was  a  man  of  the  most  singularly  blameless  life;  his  disposition 
was  amiable;  his  manner  gentle  and  unobtrusive;  and  his  decision,  when 
circumstances  demanded  it,  was  prompt  and  firm  as  the  rocks.  He  did 
nothing  from  impulse;  he  carefully  considered  his  course  and  came  to 
conclusions  which  his  conscience  approved;  and  when  his  decision  was 
made,  it  was  unalterable. 

Professor  Seitz  was  not  only  a  good  mathematician,  but  he  was  also 
proficient  in  other  branches  of  knowledge.  His  mind  was  cast  in  a  large 
mold.  ''Being  devout  in  heart  as  well  as  great  in  intellect,  'signs  and  quan- 
tities were  to  him  but  symbols  of  God's  eternal  truth'  and  'he  looked 
through  nature  up  to  nature's  God.'  Professor  Seitz,  in  the  very  appro- 
priate words  of  Dr.  Peabody  regarding  Benjamin  Pierce,  Professor  of 
Mathematics  and  Astronomy  in  Harvard  University,  'saw  things  precisely 
as  they  are  seen  by  the  infinite  mind.  He  held  the  scales  and  compasses 
with  which  eternal  wisdom  built  the  earth,  and  meted  out  the  heavens.  As 
a  mathematician,  he  was  adored  by  his  friends  with  awe.  As  a  man,  he 
was  a  Christian  in  the  whole  aim  and  tenor  of  life.'  " 

Professor  Seitz  did  not  gain  his  knowledge  from  books,  for  his  library 
consisted  of  only  a  few  books  and  periodicals.  He  gained  such  a  profound 
insight  in  the  subtle  relations  of  numbers  by  close  application  with  which 
he  was  particularly  gifted.  He  was  not  a  mathematical  genius,  that  is,  as 
ususally  understood,  one  who  is  born  with  mathematical  powers  fully  de- 
veloped. But  he  was  a  genius  in  that  he  was  especially  gifted  with  the 
power  to  concentrate  his  mind  upon  any  subject  he  wished  to  investigate. 
This  happy  faculty  of  concentrating  all  his  powers  of  mind  upon  one  topic 
to  the  exclusion  of  all  others,  and  viewing  it  from  all  sides,  enabled  him  to 
proceed  with  certainty  where  others  would  become  confused  and  disheart- 
ened. Thread  by  thread  and  step  by  step,  he  took  up  and  followed  out 
long  lines  of  thought  and  arrived  at  correct  conclusions.  The  darker  and 
more  subtle  the  question  appeared  to  the  average  mind,  the  more  eagerly 
he  investigated  it.  No  conditions  were  so  complicated  as  to  discourage 
him.  His  logic  was  overwhelming. 


442  FINKEL'S  SOLUTION  BOOK. 


BIOGRAPHY. 


RENE'  DESCARTES. 

Rene  Descartes,  the  first  of  the  modern  school  of  mathematicians, 
was  born  at  La  Haye,  a  small  town  on  the  right  bank  of  the  Creuse  and 
about  midway  between  Tours  and  Poitiers,  on  March  31st,  1596,  and 
died  at  Stockholm,  on  February  llth,  1650.  "The  house  is  still  shown 
where  he  was  born,  and  a  metairie  about  three  miles  off  still  retains  the 
name  of  Les  Cartes.  His  family  on  both  sides  was  of  Poitevin  descent 
and  had  its  headquarters  in  the  neighboring  town  of  Chatterault,  where 
his  grandfather  had  been  a  physician.  His  father,  Joachim  Descartes, 
purchased  a  commission  as  counsellor  in  the  Parlement  Rennes  and  thus 
introduced  the  family  into  that  demi-noblesse  of  the  robe  of  which,  in 
stately  isolation  between  the  bourgeoisie  and  the  high  nobility,  main- 
tained a  lofty  rank  in  the  hierarchy  of  France.  For  one-half  of  each  year 
required  for  residence  the  elder  Descartes  removed,  with  his  wife, 
Jeanne  Brochard,  to  Rennes.  Three  children,  all  of  whom  first  saw 
the  light  at  La  Haye,  sprang  from  the  union,  —  a  son,  who  afterwards 
succeeded  to  his  father  in  the  Parlement,  a  daughter  who  married  a  M. 
du  Crevis,  and  a  second  son,  Rene.  His  mother,  who  had  been  ailing  be- 
forehand, never  recovered  from  her  third  confinement;  and  the  mother- 
less infant  was  intrusted  to  a  nurse,  whose  care  Descartes  in  after  years 
remembered  by  a  small  pension."* 

Descartes,  who  early  showed  an  inquisitive  mind,  was  called  by  his 
father,  "my  philosopher."  At  the  age  of  eight,  Descartes  was  sent  to 
the  school  of  La  Fleche,  which  Henry  IV  had  lately  founded  and  en- 
dowed for  the  Jesuits,  and  here  he  continued  from  1604  to  1612.  Of 
the  education  here  given,  of  the  equality  maintained  among  the  pupils, 
and  of  their  free  intercourse,  he  spoke  at  a  later  period  in  terms  of 
high  praise.  Descartes  himself  enjoyed  exceptional  privileges.  His 
feeble  health  excused  him  from  the  morning  duties,  and  thus  early  he 
acquired  the  habit  of  matutinal  reflection  in  bed,  which  clung  to  him 
throughout  life.  When  he  visited  Pascal  in  1647,  he  told  him  that  the 
only  way  to  do  good  work  in  mathematics  and  to  preserve  his  health 
was  never  to  allow  any  one  to  make  him  get  up  in  the  morning  before 
he  felt  inclined  to  do  so.  Even  at  this  period  he  had  begun  to  distrust 
the  authority  of  tradition  and  his  teachers. 

Two  years  before  leaving  school  (1610)  he  was  selected  as  one  of 
twenty-four  gentlemen  who  went  forth  to  receive  the  heart  of  the  mur- 
dered king  as  it  was  borne  to  its  Besting  place  at  La  Fleche.  During 
the  winter  of  1612,  he  completed  his  preparations  for  the  world  by  les- 
sons in  horsemanship  and  fencing;  and  then  in  the  spring  of  1613  he 
started  for  Paris  to  be  introduced  to  the  world  of  fashion.  Fortunately 
the  spirit  of  dissipation  did  not  carry  him  very  far,  the  worst  being  a 
passion  for  gaming.  Here  through  the  medium  of  the  Jesuits  he  made 
the  acquaintance  of  Mydorge,  one  of  the  foremost  mathematicians  of 
France,  and  renewed  his  schoolboy  friendship  with  Father  Mersenne,  and 
together  with  them  he  devoted  the  two  years  of  1615  and  1616  to  the  study 
of  mathematics. 

"The  withdrawal  of  Mersenne  in  1614  to  a  post  in  the  provinces  was 
the  signal  for  Descartes  to  abandon  social  life  and  shut  himself  up  for 
nearly  two  years  in  a  secluded  house  of  the  Faubourg  St.  Germain. 
Accident,  however,  betrayed  the  secret  of  his  retirement;  he  was  com- 
pelled to  leave  his  mathematical  investigations  and  to  take  a  part  in 
entertainments,  where  the  only  thing  that  chimed  in  with  his  theorizing 

*Britannica  Encyclopedia,  Ninth  Edition. 


BIOGRAPHY.  443 

reveries  was  the  music.  The  scenes  of  horror  and  intrigue  which 
marked  the  struggle  for  supremacy  between  the  various  leaders  who 
aimed  at  guiding  the  politics  of  France  made  France  no  fit  place  for  a 
student  and  held  out  little  honorable  prospect  for  a  soldier.  Accordingly, 
in  May,  1617,  Descartes,  now  twenty-one  years  of  age,  set  out  for  the 
Netherlands,  and  took  service  in  the  army  of  Prince  Maurice  of  Orange, 
one  of  the  greatest  generals  of  the  age,  who  had  been  engaged  for  some 
time  in  a  war  with  the  Spanish  forces  in  Belgium.  At  Breda,  he  enlisted 
as  a  volunteer,  and  the  first  and  only  pay  which  he  accepted  he  kept  as  a 
curiosity  through  life.  There  was  a  lull  in  the  war;  and  the  Nether- 
lands were  distracted  by  the  quarrels  of  Gomarists  and  Arminians.  Dur- 
ing the  leisure  thus  arising,  Descartes  one  day,  as  he  roved  through 
Breda,  had  his  attention  drawn  to  a  placard  in  the  Dutch  tongue; 
and  as  the  language  of  which  he  never  became  perfectly  master,  was 
then  strange  to  him,  he  asked  a  bystander  to  interpret  it  in  either  French 
or  Latin.  The  stranger,  who  happened  to  be  Isaac  Beeckman,  principal 
of  the  College  of  Dort,  offered  with  some  surprise  to  do  so  into  Latin, 
if  the  inquirer  would  bring  him  a  solution  of  the  problem  —  for  the 
advertisement  was  one  of  those  challenges  which  the  mathematicians  of 
the  age,  in  the  spirit  of  the  tournament  of  chivalry,  were  accustomed  to 
throw  down  to  all  comers,  daring  them  to  discover  a  geometrical  mystery 
known  as  they  fancied  to  themselves  alone.  Descartes  promised  and  ful- 
filled ;  and  a  friendship  grew  up  between  him  and  Beeckman  —  broken 
only  by  the  literary  dishonesty  of  the  latter,  who  in  later  years  took 
credit  for  the  novelty  contained  in  a  small  essay  on  music  (Compendium 
Musicae)  which  Descartes  wrote  at  this  period  and  intrusted  to 
Beeckman."* 

The  unexpected  test  of  his  mathematical  attainments  afforded  by  the 
solution  of  the  problem  referred  to,  its  solution  costing  him  only  a  few 
liours  study,  made  the  uncongenial  army  life  distasteful  to  him,  but  under 
family  influence  and  tradition,  he  remained  a  soldier,  and  was  pursuaded 
at  the  commencement  of  the  thirty  years'  war  to  volunteer  under  Count 
de  Bucquoy  in  the  army  of  Bavaria.  The  winter  of  1619,  spent  in 

Siarters  at  Neuburg  on  the  Danube,  was  the  critical  period  in  his  life, 
ere,  in  his  warm  room  (dans  un  poele),  he  indulged  those  meditations 
which  afterwards  led  to  the  Discours  de  la  Methode  (Discourse  of 
Method).  It  was  here  that,  on  the  eve  of  St.  Martin's  day,  November 
10,  1619,  he  "was  filled  with  enthusiasm,  and  discovered  the  foundations 
of  a  marvelous  science." 

He  retired  to  rest  with  anxious  thoughts  of  his  future  career,  which 
haunted  him  through  the  night  in  three  dreams,  that  left  deep  impres- 
sions on  his  mind.  "Next  day,"  he  says,  "I  began  to  understand  the 
first  principles  of  my  marvelous  discovery."  Thus  the  date  of  his  philo- 
sophical conversion  is  fixed  to  a  day.  This  day  marks  the  birth  of 
modern  mathematics.  His  discovery,  viz.,  the  cooperation  of  ancient 
geometry  and  algebra,  is  epoch-making  in  the  history  of  mathematics. 

It  is  frequently  stated  that  Descartes  was  the  first  to  apply  algebra 
to  geometry.  This  statement  is  not  true,  for  Vieta  and  others  had  done 
this  before  him,  and  even  the  Arabs  sometimes  used  algebra  in  connec- 
tion with  geometry.  "The  new  step  that  Descartes  did  take  was  the  in- 
troduction into  geometry  of  an  analytical  method  based  on  the  notion 
of  variables  and  constants,  which  enabled  him  to  represent  curves  by 
algebraic  equations.  In  the  Greek  geometry,  the  idea  of  motion  was 
wanting,  but  with  Descartes  it  became  a  very  fruitful  conception.  By 
him  a  point  was  determined  in  position  by  its  distances  from  two  fixed 
lines  or  axes.  These  distances  varied  with  every  change  of  position  in 
the  point.  This  geometric  idea  of  co-ordinate  representation  together  with 
the  algebraic  idea  of  two  'variables  in  one  equation  having  an  indefinite 
number  of  simultaneous  values,  furnished  a  method  for  the  study  of  loci, 

"••Encyclopedia  Britannica,  Ninth  Edition. 


444  FINKEL'S  SOLUTION  BOOK. 

which  is  admirable  for  the  generality  of  its  solutions.  Thus  the  entire 
conic  sections  of  Appollonius  is  wrapped  up  and  contained  in  a  single 
equation  of  the  second  degree."f 

''Descartes  found  in  mathematics,  as  did  Kant  and  Comte,  the  type 
of  all  faultless  thought;  and  he  oroved  his  appreciation  of  his  insighf 
by  the  invention  of  a  new  symbolic  mechanism  and  artifice  for  the  appli- 
cations of  algebra  to  geometry  (Analytic  Geometry,  as  it  is  now  calledj 
which,  in  a  growing  sense,  let  it  be  said,  existed  before  him),  and  by 
his  discoveries  in  the  theory  of  equations,  which  were  fundamental  in 
their  importance."* 

After  a  short  sojourn  in  Paris,  Descartes  moved  to  Holland,  then  at 
the  height  of  its  power.  There  for  twenty  years  he  lived,  giving  up  all 
his  time  to  philosophy  and  mathematics.  Science,  he  says,  may  be 
compared  to  a  tree,  metaphysics  is  the  root,  physics  is  the  trunk,  and 
the  three  chief  branches  are  mechanics,  medicine,  and  morals,  these 
forming  the  three  applications  of  our  knowledge,  namely,  to  the  external 
world,  to  the  human  body,  and  to  the  conduct  of  life;  and  with  these 
subjects  alone  his  writings  are  concerned. 

He  spent  the  time  from  1629  to  1633  writing  Le  Monde,  a  work  em- 
bodying an  attempt  to  give  a  physical  theory  of  the  universe;  but  find- 
ing its  publication  likely  to  bring  on  him  the  hostility  of  the  Church, 
and  having  no  desire  to  pose  as  a  martyr,  he  abandoned  it.  The  in- 
complete manuscript  was  published  in  1664. 

He  then  devoted  himself  to  composing  a  treatise  on  universal  science ; 
this  was  published  at  Leyden  in  1637  under  the  title  Discourse  de  la 
methode  pour  bien  conduire  sa  raison  et  chercher  la  verite  dans  les 
sciences,  and  was  accompanied  with  three  appendices  entitled  La  Diop- 
trique,  Les  Meleores,  and  La  Geometric.  It  is  from  the  last  of  these 
that  the  invention  of  analytical  geometry  dates.  In  1641,  he  published  a 
work  called  Meditations,  in  which  he  explained  at  some  length  his 
views  of  philosophy  as  sketched  out  in  the  Discourse.  In  1644,  he  issued 
the  Principia  Philosophiae,  the  greater  part  of  which  was  devoted  to 
physical  science  especially  the  laws  of  motion  and  the  theory  of  vor- 
tices. In  his  theory  of  vortices,  he  commences  with  a  discussion  of 
motion ;  and  then  lays  down  ten  laws  of  nature,  of  which  the  first 
two  are  almost  identical  with  the  first  two  as  laid  down  by  Newton. 
The  remaining  eight  are  inaccurate.  He  next  proceeds  to  a  discussion 
of  the  nature  of  matter  which  he  regards  uniform  in  kind  though  there 
are  three  forms  of  it.  He  assumes  that  the  matter  of  the  universe  is  in 
motion,  that  this  motion  is  constant  in  amount,  and  that  the  motion 
results  in  a  number  of  vortices.  He  states  that  the  sun  is  the  center  of 
an  immense  whirlpool  of  this  matter,  in  which  the  planets  float  and 
are  swept  round  like  straws  in  a  whirlpool  of  water. 

Each  planet  is  supposed  to  be  the  center  of  a  secondary  whirlpool  by 
which  its  satellites  are  carried,  and  so  on.  All  of  these  assumptions  are 
arbitrary  and  unsupported  by  any  investigation.  It  is  a  little  strange 
that  a  man  who  began  his  philosophical  reasonings  by  doubting  all 
things  and  finally  coming  to  cogito,  ergo  sum  should  have  made  assump- 
tions so  groundless. 

While  Descartes  was  a  philosopher  of  a  very  high  type,  yet  his  fame 
will  ever  rest  on  his  researches  in  mathematics.  The  first  important 
problem  solved  by  Descartes  in  his  geometry  is  the  problem  of  Pappus, 
viz. :  "Given  several  straight  lines  in  a  plane,  to  find  the  locus  of  a  point 
such  that  perpendiculars,  or,  more  generally,  straight  lines  at  given  angles, 
drawn  from  the  point  to  the  given  lines,  shall  satisfy  that  the  product  of 
certain  of  them  shall  be  in  given  ratio  to  the  product  of  the  rest."  "The 
most  important  case  of  this  problem  is  to  find  the  locus  of  a  point  such 
that  the  product  of  the  perpendiculars  on  m  given  lines  be  in  a  constant 

•\Cajori' s  History  of  Mathematics. 
*The  Open  Court,  August,  1898. 


BIOGRAPHY.  445 

ratio  to  the  product  of  the  perpendiculars  on  n  other  given  straight 
lines.  The  ancients  had  solved  this  geometrically  for  the  case  m  =  1, 
n  =  1,  and  the  case  w  =  1,  n  =  2.  Pappus  had  further  stated  that  if 
w  — n  — 2,  the  locus  was  a  conic,  but  he  gave  no  proof.  Descartes  also 
failed  to  prove  this  by  pure  geometry,  but  he  showed  that  the  curve  was 
represented  by  an  equation  of  the  second  degree,  that  is,  was  a  conic; 
subsequently  Newton  gave  an  elegant  solution  of  the  problem  by  pure 
geometry."* 

In  algebra,  Descartes  expounded  and  illustrated  the  general  methods 
of  solving  equations  up  to  those  of  the  fourth  degree  (and  believed  that 
his  method  could  go  beyond),  stated  the  law  which  connects  the  posi- 
tive and  negative  roots  of  an  equation  with  the  change  of  signs  in  the 
consecutive  terms,  known  as  Descartes'  Law  of  Signs,  and  introduced 
the  method  of  indeterminate  coefficients  for  the  solution  of  equations. 

In  appearance,  Descartes  was  a  small  man  with  large  head,  project- 
ing brow,  prominent  nose,  and  black  hair  coming  down  to  his  eye- 
brows. His  voice  was  feeble.  Considering  the  range  of  his  studies  he 
was  by  no  means  widely  read,  had  no  use  for  Greek,  as  is  shown  by 
his  disgust  when  he  found  that  Queen  Christina  devoted  some  time  each 
day  to  its  study,  and  despised  both  learning  and  art  unless  something 
tangible  could  be  extracted  therefrom.  In  philosophy,  he  did  not  read 
much  of  the  writings  of  others.  In  disposition,  he  was  cold  and  selfish. 
He  never  married,  and  left  no  descendants,  though  he  had  one  illegitimate 
daughter,  Francine,  who  died  in  1640,  at  the  age  of  five. 

In  1649,  through  the  instigation  of  his  close  personal  friend,  Chanut, 
he  received  an  invitation  to  the  Swedish  court,  and  in  September  of  that 
year  he  left  Egmond  for  the  north.  Here,  on  the  llth  of  February,  1650, 
he  died  of  inflammation  of  the  lungs  brought  about  by  too  close  devo- 
tion to  the  sick-room  of  his  friend  Chanut,  who  was  dangerously  ill 
with  the  same  disease.  —  By  B.  F.  Finkel.  From  the  American  Mathe- 
matical Monthly. 

*  Ball's  Short  Account  of  the  History  of  Mathematics. 


446  FINKEIv'S  SOLUTION  BOOK. 


BIOGRAPHY. 


LEONHARD  EULER. 

Leonhard  Euler  (oiler),  one  of  the  greatest  and  most  prolific  mathe- 
maticians that  the  world  has  produced,  was  born  at  Basel,  Switzerland, 
on  the  15th  day  of  April,  1707,  and  died  at  St.  Petersburg,  Russia,  No- 
vember the  18th  (N.  S.),  1783.  Euler  received  his  preliminary  instruc- 
tion in  mathematics  from  his  father  who  had  considerable  attainments  as 
a  mathematician,  and  who  was  a  Calvinistic*  pastor  of  the  village  of 
Riechen,  which  is  not  far  from  Basel.  He  was  then  sent  to  the  Univer- 
sity of  Basel  where'  he  studied  mathematics  under  the  direction  of  John 
Bernoulli,  with  whose  two  sons,  Daniel  and  Nicholas,  he  formed  a  life- 
long friendship.  Geometry  soon  became  his  favorite  study.  His  genius 
for  analytical  science  soon  gained  for  him  a  high  place  in  the  esteem 
of  his  instructor,  John  Bernoulli,  who  was  at  the  time  one  of  the  first 
mathematicians  of  Europe.  Having  taken  his  degree  as  Master  of  Arts 
in  1723,  Euler  afterwards  applied  himself,  at  his  father's  desire,  to  the 
study  of  theology  and  the  Oriental  languages,  with  the  view  of  entering 
the  ministry,  but,  with  his  father's  consent,  he  returned  to  his  favorite 
pursuit,  the  study  of  mathematics.  At  the  same  time,  by  the  advice  of 
the  younger  Bernoulli  is,  who  had  removed  to  St.  Petersburg  in  1725, 
he  applied  himself  to  the  study  of  physiology,  to  which  he  made  useful 
applications  of  his  mathematical  knowledge ;  he  also  attended  the  lectures 
of  the  most  eminent  professors  of  Basel.  While  he  was  eagerly  engaged 
in  physiological  researches,  he  composed  a  dissertation  on  the  nature 
and  propagation  of  sound.  In  his  nineteenth  year  he  also  composed  a 
dissertation  in  answer  to  a  prize-question  concerning  the  masting  of  ships, 
for  which  he  received  the  second  prize  from  the  French  Academy  of 
Sciences. 

When  his  two  close  friends,  Daniel  and  Nicholas  Bernoulli,  went  to 
Russia,  they  induced  Catherine  I,  in  1727,  to  invite  Euler  to  St.  Peters- 
burg, where  Daniel,  in  1733,  was  assigned  to  the  chair  of  mathematics. 
Euler  took  up  his  residence  in  St.  Petersburg,  and  was  made  an  asso- 
ciate of  the  Academy  of  Sciences.  In  1730  he  became  professor  of  physics, 
and  in  1733  he  succeeded  his  friend  Daniel  Bernoulli,  who  resigned  on  a 
plea  of  ill  health. 

At  the  commencement  of  his  astonishing  career,  he  enriched  the 
Academical  collection  with  many  memoirs,  which  excited  a  noble  emula- 
tion between  him  and  the  Bernouillis,  though  this  did  not  in"  any  way 
affect  their  friendship.  It  was  at  this  time  that  he  carried  the  integral 
calculus  to  a  higher  degree  of  perfection,  invented  the  calculation  of 
sines,  reduced  analytical  operations  to  greater  simplicity,  and  threw  new 
light  on  nearly  all  parts  of  pure  or  abstract  mathematics.  In  1735,  an 
astronomical  problem  proposed  by  the  Academy,  for  the  solution  gi  which 
several  eminent  mathematicians  had  demanded  several  months'  time, 


*The  Encyclopedia  Britannica  says  Ruler's  father  was  a  Calvinistic  minister,  while 
W.  W.  R.  Ball,  in  his  History  of  Mathematics,  says  he  was  a  Lutheran  minister.  Euler 
himself  was  a  Calvinist  in  doctrine,  as  the  following,  which  is  his  apology  for  prayer, 
indicates:  "I  remark,  first,  that  when  God  established  the  course  of  the  universe,  and 
arranged  all  the  events  which  must  come  to  pass  in  it,  he  paid  attention  to  all  the  cir- 
cumstances which  should  accompany  each  event ;  and  particularly  to  the  dispositions, 
to  the  desires,  and  prayers  of  every  intelligent  being;  and  that  the  arrangement  of  all 
events  was  disposed  in  perfect  harmony  with  all  these  circumstances.  When,  therefore, 
a  man  addresses  God  a  prayer  worthy  of  being  heard  it  must  not  be  imagined  that  such 
a  prayer  came  not  to  the  knowledge  of  God  till  the  moment  it  was  formed.  That  prayer 
was  already  heard  from  all  eternity;  and  if  the  Father  of  Mercies  deemed  it  worthy 
of  being  answered,  he  arranged  the  world  expressly  in  favor  of  that  prayer,  so  that  the 
accomplishment  should  be  a  consequence  of  the  natural  course  of  events.  It  is  thus 
that  God  answers  the  prayers  of  men  without  working  a  miracle." 


BIOGRAPHY.  447 

was  solved  by  Euler  in  three  days  with  the  aid  of  improved  methods  of 
his  own,  but  the  effort  threw  him  into  a  fever  which  endangered  his  life 
and  deprived  him  of  his  right  eye,  his  eyesight  having  been  impaired 
by  the  severity  of  the  climate.  With  still  superior  methods,  this  same 
problem  was  solved  later  by  the  illustrious  German  mathematician, 
Gauss. 

In  1741,  at  the  request,  or  rather  command,  of  Frederick  the  Great, 
he  moved  to  Berlin,  where  he  was  made  a  member  of  the  Academy  of 
Sciences,  and  Professor  of  Mathematics.  He  enriched  the  last  volume  of 
the  Melanges  or  Miscellanies  of  Berlin,  with  five  memoirs,  and  these  were 
followed,  with  astonishing  rapidity,  by  a  great  number  of  important  re- 
searches, which  wefce  scattered  throughout  the  annual  memoirs  of  the 
Prussian  Academy.  At  the  same  time,  he  continued  his  philosophical 
contributions  to  the  Academy  of  St.  Petersburg,  which  granted  him  a 
pension  in  1742. 

The  respect  in  which  he  was  held  by  the  Russians  was  strikingly 
shown  in  1760,  when  a  farm  he  occupied  near  Charlottenburg  happened 
to  be  pillaged  by  the  invading  Russian  army.  On  its  being  ascertained 
that  the  farm  belonged  to  Euler,  the  general  immediately  ordered  com- 
pensation to  be  paid,  and  the  Empress  Elizabeth  sent-  an  'additional  sum 
of  four  thousand  crowns.  The  despotism  of  Anne  I  caused  Euler,  who 
was  a  very  timid  man,  to  shrink  from  public  affairs,  .and  to  devote  all 
his  time  to  science.  After  his  call  to  Berlin,  the  Queen  of  Prussia  who 
received  him  kindly,  wondered  how  so  distinguished  a  scholar  should  be 
so  timid  and  reticent.  Euler  replied,  "Madam,  it  is  because  I  come  from 
a  country  where,  when  one  speaks,  one  is  hanged." 

In  1766,  Euler,  with  difficulty,  obtained  permission  from  the  King  of 
Prussia  to  return  to  St.  Petersburg,  to  which  he  had  been  originally 
called  by  Catherine  II.  Soon  after  returning  to  St.  Petersburg  a  cataract 
formed  in  his  left  eye,  which  ultimately  -deprived  him  of  sight,  but  this 
did  not  stop  his  wonderful  literary  productiveness,  which  continued  for 
seventeen  years  —  until  the  day  of  his  death.  It  was  under  these  cir- 
cumstances that  he  dictated  to  his  amanuensis,  a  tailor's  apprentice  who 
was  absolutely  devoid  of  mathematical  knowledge,  his  Anleitung  zur 
Algebra,  or  Elements  of  Algebra,  1770,  a  work  which,  though  purely 
elementary,  displays  the  mathematical  genius  of  its  author,  and  is  still 
considered  one  of  the  best  works  of  its  class.  Euler  was  one  of  the  very 
few  great  mathematicians  who  did  not  deem  it  beneath  the  dignity  of 
genius  to  give  some  attention  to  the  recasting  of  elementary  processes 
and  the  perfecting  of  elementary  text-books,  and  it  is  not  improbable 
that  modern  mathematics  is  as  greatly  indebted  to  him  for  his  work 
along  this  line  as  for  his  original  creative  work. 

Another  task  to  which  he  set  himself  soon  after  returning  to  St. 
Petersburg  was  the  preparation  of  his  Lettres  a  une  Princesse  d'  Allemagne 
sur  quelques  sujects  de  Physique,  (3  yols.  1768-72).  These  letters  were 
written  at  the  request  of  the  princess  of  Anhalt-Dessau,  and  contain  an 
admirably  clear  exposition  of  the  principal  facts  of  mechanics,  optics, 
acoustics,  and  physical  astronomy.  Theory,  however,  is  frequently  un- 
soundly applied  in  it,  and  it  is  to  be  observed  generally  that  Euler's 
strength  lay  rather  in  pure  than  in  applied  mathematics.  In  1755,  Euler 
had  been  elected  a  foreign  member  of  the  Academy  of  Sciences  at  Paris, 
and  sometime  afterwards  the  academical  prize  was  adjudged  to  three  of 
his  memoirs  Concerning  the  Inequalities  in  the  Motions  of  the  Planets. 
The  two  prize-problems  proposed  by  the  same  Academy  in  1770  and 
1772  were  designed  to  obtain  a  more  perfect  theory  of  the  moon's  mo- 
tion. Euler,  assisted  by  his  eldest  son,  Johann  Albert,  was  a  competitor 
for  these  prizes  and  obtained  both.  In  his  second  memoir,  he  reserved 
for  further  consideration  the  several  inequalities  of  the  moon's  motion, 
which  he  could  not  determine  in  riis  first  theory  on  account  9f  the 
complicated  calculations  in  which  the  method  he  then  employed  had 


448  FINKEVS  SOLUTION  BOOK. 

engaged  him.  He  afterward  reviewed  his  whole  theory  with  the  assist- 
ance of  his  son  and  Krafft  and  Lexell,  and  pursued  his  researches  until 
he  had  consiructed  the  new  tables,  which  appeared  with  the  great  work 
in  1772.  Instead  of  confining  himself,  as  before,  to  the  fruitless  integra- 
tion of  three  differential  equations  of  the  second  degree,  which  are  fur- 
nished by  mathematical  principles,  he  reduced  them  to  three  ordinates 
which  determine  the  place  of  the  moon ;  and  he  divides  into  classes  all 
the  inequalities  of  that  planet,  as  far  as  they  depend  either  on  the  elonga- 
tion of  the  sun  and  moon,  or  upon  the  eccentricity,  or  the  parallax,  or 
the  inclination  of  the  lunar  orbit.  The  inherent  difficulties  of  this  task 
were  immensely  enhanced  by  the  fact  that  Euler  was  virtually  blind,  and 
had  to  carry  all  the  elaborate  computations  involved  in  his  memory.  A 
further  difficulty  arose  from  the  burning  of  his  house  and  the  destruction 
of  a  greater  part  of  his  property  in  1771.  His  manuscripts  were  fortu- 
nately preserved.  His  own  life  only  was  saved  by  the  courage  of  a 
native  of  Basel,  Peter  Grimmon,  who  carried  him  out  of  the  burning 
house. 

Some  time  after  this,  the  celebrated  Wenzell,  by  couching  the  cataract, 
restored  his  sight ;  but  a  too  harsh  use  of  the  recovered  faculty,  together 
with  some  carelessness  on  the  part  of  the  surgeons,  brdught  about  a 
relapse.  With  the  assistance  of  his  sons,  and  of  Krafft  and  Lexell,  how- 
ever, he  continued  his  labors,  neither  the  loss  of  his  sight  nor  the  in- 
firmities of  an  advanced  age  being  sufficient  to  check  his  activity.  .Hav- 
ing engaged  to  furnish  the  Academy  of  St.  Petersburg  with  as  many 
memoirs  as  would  be  sufficient  to  complete  its  acts  for  twenty  years  after 
his  death,  he  in  seven  years  transmitted  to  the  Academy  above  seventy 
memoirs,  and  left  above  two  hundred  more,  which  were  revised  and  com- 
pleted by  another  hand. 

Euler's  knowledge  was  more  general  than  might  have  been  expected 
in  one  who  had  pursued  with  such  unremitting  ardor,  mathematics  and 
astronomy,  as  his  favorite  studies.  He  had  made  considerable  progress 
in  medicine,  botany,  and  chemistry,  and  he  was  an  excellent  classical 
scholar  and  extensively  read  in  general  literature.  He  could  repeat  the 
Aenied  of  Virgil  from  the  beginning  to  the  end  without  hesitation,  and 
indicate  the  first  and  last  line  of  every  page  of  the  edition  which  he  used. 
But  such  lines  from  Virgil  as,  "The  anchor  drops,  the  rushing  keel  is 
staid,"  always  suggested  to  him  a  problem  and  he  could  not  help  en- 
quiring what  would  be  the  ship's  motion  in  such  a  case. 

Euler's  constitution  was  uncommonly  vigorous  and  his  general  health 
was  always  good.  He  was  enabled  to  continue  his  labors  to  the  very  close 
of  his  life  so  that  it  was  said  of  him,  that  he  ceased  to  calculate  and  to 
breathe  at  nearly  the  same  moment.  His  last  subject  of  investigation 
was  the  motions  of  balloons,  and  the  last  subject  on  which  he  conversed 
was  the  newly  discovered  planet  Herschel. 

On  the  18th  of  September,  1783,  while  he  was  amusing  himself  at  tea 
with  one  of  his  grandchildren,  he  was  struck  with  apoplexy,  which  ter- 
minated the  illustrious  career  of  this  wonderful  genius,  at  the  age  of 
seventy-six.  His  works,  if  printed  in  their  completeness,  would  occupy 
from  60  to  80  quarto  volumes.  However,  no  complete  edition  of  Euler's 
writings  has  been  published,  though  the  work  has  been  begun  twice. 

He  was  simple,,  upright,  affectionate,  and  had  a  strong  religious  faith. 
His  single  and  unselfish  devotion  to  the  truth  and  his  joy  at  the  dis- 
coveries of  science  whether  made  by  himself  or  others,  were  striking  attri- 
butes of  his  character.  He  was  twice  married,  his  second  wife  being  a 
half-sister  of  his  first,  and  he  had  a  numerous  family,  several  of  whom 
attained  to  distinction.  His  eloge  was  written  for  the  French  Academy 
by  Condorcet,  and  an  account  of  his  life,  with  a  list  of  his  works,  was 
written  by  Von  Fuss,  the  secretary  of  the  Imperial  Academy  of  St. 
Petersburg. 


BIOGRAPHY. 


449 


As  has  been  said,  Euler  wrote  an  immense  number  of  works,  chief 
of  which  are  the  following:  Introductio  in  Analysin  infinitorum,  1748, 
which  was  intended  to  serve  as  an  introduction  to  pure  analytical  mathe- 
matics. This  work  produced  a  revolution  in  analytical  mathematics,  as 
the  subject  of  which  it  treated  had  hitherto  never  been  presented  in  so 
general  and  systematic  a  manner.  The  first  part  of  the  Analysis  Infini- 
torum contains  the  bulk  of  the  matter  which  is  to  be  found  in  modern 
text-books  on  algebra,  theory  of  equations,  and  trigonometry.  In  the 
algebra,  he  paid  particular  attention  to  the  expansion  of  various  functions 
in  series,  and  to  the  summation  of  given  series,  and  pointed  out  explicitly 
that  an  infinite  series  can  not  be  safely  employed  in  mathematical  inves- 
tigations unless  it  is  convergent.  In  trigonometry,  he  introduced  (simul- 
taneously with  Thomas  Simpson  in  England)  the  now  current  abbrevia- 
tions for  trigonometric  functions,  and  simplified  formulas  by  the  simple 
expedient  of  designating  the  angles  of  a  triangle  by  A,  B,  C,  and  the  op- 
posite sides  by  a,  b,  c.  He  also  showed  that  the  trigonometrical  and  ex- 
ponential functions  are  connected  by  the  relation  cos#-Hsin#=^'#.  Here 
too  we  meet  the  symbol  e  used  to  denote  the  base  of  the  Naperian 
logarithms,  namely  the  incommensurable  number  2.7182818  .  .  .  and 
the  symbol  ?r  used  to  denote  the  incommensurable  number  3.14159265  . 
.  .  The  use  of  a  single  symbol  to  denote  the  number  2.7182818  .  .  . 
•seems  to  be  due  to  Cotes,  who  denoted  it  by  M.  Newton  was  probably 
the  first  to  employ  the  literal  exponential  notation,  and  Euler  using  the 
form  a%,  had  taken  a  as  the  base  of  any  system  of  logarithms.  It  is 
probable  that  the  choice  of  e  for  a  particular  base  was  determined  by 
its  being  the  vowel  consecutive  to  a,  or,  still  more  probable  because  e 
is  the  initial  of  the  word  exponent. 

The  use  of  a  single  symbol  to  denote  3.14159265  .  .  .  appears  to 
have  been  introduced  by  John  Bournilli,  who  represented  it  by  c.  Euler 
in  1734  denoted  it  by  p,  and  in  a  letter  of  1736  in  which  he  enunciated  the 
theorem  that  the  sum  of  the  square  of  the  reciprocals  of  the  natural 
numbers  is  ^-Tr2,  he  uses  the  letter  c.  The  symbol  TT  was  first  used 
to  represent  3.141592  ...  by  William  Jones's  in  his  "Synopsis  Pal- 
mariorum  Matheseos",  London,  1706,  and  after  the  publication  of  Euler's 
Analysis,  the  symbol  TT  was  generally  employed,  the  choice  of  TT  being 
determined  by  the  initial  of  the  word,  Tt£pi<p£  peta  =  periphereia. 

The  second  part  of  the  Analysis  Infinitorum  is  on  analytical  geometry. 
Euler  begins  this  part  by  dividing  curves  into  algebraic  and  transcendental, 
and  establishes  a  number  of  propositions  which  are  true  for  all  algebraic 
curves.  He  then  applied  these  to  the  general  equation  of  the  second 
degree  in  two  dimensions,  showed  that  it  represents  the  various  conic 
sections,  and  deduces  most  of  their  properties  from  the  general 'equation. 
He  also  considered  the  classification  of  cubic,  quartic,  and  other  alge- 
braic curves.  He  next  discussed  the  question  as  to  what  surfaces  are 
represented  by  the  general  equation  of  the  second  degree  in  three  dimen- 
sions, and  how  they  may  be  discriminated  one  from  the  other.  Some 
of  these  surfaces  had  not  been  previously  investigated.  In  this  work  he 
also  laid  down  the  rules  for  the  transformation  of  coordinates  in  space. 
Here  also  we  find  the  first  attempt  to  bring  the  curvature  of  surfaces 
within  the  domain  of  mathematics,  and  the  first  complete  discussion  of 
tortuous  curves. 

In  1755  appeared  Institutions  Calculi  Differential,  to  ^which  the 
Analysis  Infinitorum  was  intended  as  an  introduction.  This  is  the  first 
text-book  on  the  differential  calculus  which  has  any  claim  to  be  regarded 
as  complete,  and  it  may  be  said  that  most  modern  treatises  on  the  sub- 
ject are  based  upon  it. 

At  the  same  time,  the  exposition  of  the  principles  of  the  subject  is 
often  prolix  and  obscure,  and  sometimes  not  quite  accurate. 


450  FINKEVS  SOLUTION  BOOK. 

This  series  of  works  was  completed  by  the  publication  in  three  vol- 
umes in  1768  to  1770  of  the  Institutiones  Calculi  Integralis,  in  which  the 
results  of  several  of  Euler's  earlier  memoirs  on  the  same  subjects  and 
on  differential  equations  are  included.  In  this  treatise  as  in  the  one  on. 
the  differential  calculus  was  summed  up  all  that  was  at  that  time  known 
on  the  subject.  The  beta  and  gamma  functions  were  invented  by  Euler, 
and  are  discussed  here,  but  only  as  methods  of  reduction  and  integration. 
His  treatment  of  elliptic  integrals  is  superficial.  The  classic  problems  on. 
isoperimetrical  curves,  the  brachistochrone  in  a  resisting  medium,  and 
theory  of  geodesies  had  engaged  Euler's  attention  at  an  early  date,  and 
the  solving  of  which  led  him  to  the  calculus  of  variations.  The  general 
idea  of  this  was  laid  down  in  his  Curvarum  Maximi  Minimise  Proprietate 
Gaudentium  Inventio  Nova  ac  Facilis,  published  in  1744,  but  the  com- 
plete development  of  the  new  calculus  was  first  effected  by  Lagrange  in 
1759.  The  method  used  by  Lagrange  is  described  in  Euler's  integral 
calculus,  and  is  the  same  as  that  given  in  most  modern  text-books  on 
the  subject. 

In  1770,  Euler  published  the  Anleitung  zur  Algebra  in  two  volumes. 
The  first  volume  treats  of  determinate  algebra.  This  work  includes  the 
proof  of  the  binomial  theorem  for  any  index,  which  is  still  known  by 
Euler's  name.  The  proof,  which  is  not  accurate  according  to  the 
modern  views  of  infinite  series,  depends  upon  the  principle  of  the  per- 
manence of  equivalent  forms,  and  may  be  seen  in  C.  Smith's  Treatise  on 
Algebra,  pages  336-7.  Euler's  proof  with  important  additions  due  to 
Cauchy,  may  be  seen  in  G.  Chrystal's  Algebra,  Part  II. 

It  is  a  fact  worthy  of  note  that  Euler  made  no  attempt  to  investigate 
the  convergency  of  the  series,  though  he  clearly  recognized  the  necessity  of 
considering  the  convergency  of  infinite  series.  While  Euler  recognized  the 
convergency  of  series,  his  conclusions  in  reference  to  infinite  series  are  not. 
always  sound.  In  his  time  no  clear  notion  as  to  what  constitutes  a  con- 
vergent series  existed,  and  the  rigid  treatment  to  which  infinite  series  are 
now  subjected  was  undreamed  of.  Euler  concluded  that  the  sum  of  the 
oscillating  series  1  —  1  +  1  —  14-1  —  1+  .  .  .  =  £,  for  the  reason,  that 
by  stopping  with  an  even  number  of  terms  the  sum  is  0,  and  by  stopping 
with  an  odd  number  of  terms  the  sum  is  1.  Hence,  the  sum  of  the  series 
is  £(0  +  1)  =  £.  Guido  Grandi  went  so  far  as  to  conclude  that  £  =  0  + 
0  +  0  +  0  .  .  .  The  paper  in  which  Euler  cautions  against  divergent 

series  contains  the  proof  that  .  .  .  — y  + 1-  1  +  n  +  «2  +  n3  .  .  .  =  0. 

His  proof  is  as  follows,  n  +  n2  +  «3  +  .  .  .  — 1  -I 1 h  •  •  * 

1  —  n  n         n* 

= :-, : — |-  ~ —  —  =  0.     Euler  had  no  hesitation  in  writing  1  —  3 

"ft  —  JL      %  —  J.          JL  —  72 

_j_5_7  +  9  — ...  =  0,  and  he  confidently  believed  that  sin^  —  2sin2^  + 
3sin3y?  —  .  .  .  =  0. 

A  remarkable  development,  due  to  Euler,  is  what  he  named  the  hyper- 
geometrical  series,  the  summation  of  which  he  observed  to  be  dependent 
upon  the  integration  of  linear  differential  equations  of  the  second  order, 
but  it  remained  for  Gauss  to  point  out  that  for  special  values  of  the 
letters,  this  series  represented  nearly  all  the  functions  then  known.  By 
giving  the  factors  641  X  6700417  of  the  number  2^  +  1  —  4294967297 
when  n  =  5,  he  pointed  out  the  fact  that  this  expression  did  not  always 
represent  primes,  as  was  supposed  by  Fermat.  —  By  B.  F.  Finkel.  From, 
the  American  Mathematical  Monthly,  Vol.  IV,  No.  12. 


BIOGRAPHY. 


451 


SOPHUS   LIE.* 

Sophus  Lie  was  born  on  the  17th  of  December,  1842,  at  Nordfjordeid 
(near  Floro)  where  his  father,  John  Herman  Lie,  was  pastor.  The 
studies  of  his  childhood  and  youth  did  not  reveal  in  him  that  exceptional 
aptitude  for  mathematics  which  is  signalized  so  early  in  the  lives  of  the 
great  geometers:  Gauss,  Abel,  and  many  others.  Even  on  leaving  the 
University  of  Christiania  in  1865,  he  still  hesitated  between  philology  and 
mathematics.  It  was  the  works  of  Pliicker  on  modern  geometry  which 
first  made  him  fully  conscious  of  his  mathematical  abilities  and  awak- 
ened within  him  an  ardent  desire  to  consecrate  himself  to  mathematical 
research.  Surmounting  all  difficulties  and  working  with  indomitable 
energy  he  published  his  first  work  in  1869,  and  we  can  say  that  from 
1870  on  he  was  in  possession  of  the  ideas  which  were  to  direct  his 
whole  career. 

At  this  time  I  frequently  had  the  pleasure  of  meeting  and  conversing 
with  him  in  Paris  where  he  had  come  with  his  friend  F.  Klein.  A  course 
of  lectures  by  Sylow  revealed  to  Lie  all  the  importance  of  the  theory  of 
substitution  groups ;  the  two  friends  studied  this  theory  in  the  great 
treatise  of  our  colleague  Jordan ;  they  saw  fully  the  essential  role  which 
it  would  be  called  upon  to  play  in  all  the  branches  of  mathematics  to 
which  it  had  then  not  been  applied.  They  have  both  had  the  good 
fortune  to  contribute  by  their  works  to  impressing  upon  mathematical 
studies  the  direction  which  appeared  to  them  to  be  the  best. 

A  short  note  of  Lie  "Sur  une  transformation  geometrique,"  pre- 
sented to  our  Academy  in  October,  1870,  contains  an  extremely  original 
discovery.  Nothing  resembles  a  sphere  less  than  a  straight  line  and 
yet,  by  using  the  ideas  of  Pliicker,  Lie  found  a  singular  transformation 
which  makes  a  sphere  correspond  to  a  straight  line,  and  which  conse- 
quently makes  possible  the  derivation  of  a  theorem  relative  to  an  en- 
semble of  spheres  from  every  theorem  relative  to  an  aggregate  of  straight 
lines,  and  vice  versa.  It  is  true  that  if  the  lines  are  real,  the  correspond- 
ing spheres  are  imaginary.  But  such  difficulties  are  not  sufficient  to 
deter  geometers.  In  this  curious  method  of  transformation,  each  property 
relative  to  asymptotic  lines  of  a  surface  is  transformed  into  a  property 
relative  to  lines  of  curvature.  The  name  of  Lie  will  remain  attached 
to  these  concealed  relations  which  connect  the  two  essential  and  funda- 
mental elements  of  geometric  investigation,  the  straight  line  and  sphere. 
He  has  developed  them  in  detail  in  a  memoir  full  of  new  ideas  which 
appeared  in  1872  in  the  Mathematische.  Annalen. 

The  works  following  this  brilliant  beginning  fully  confirmed  all  the 
hopes  to  which  it  gave  birth.  Since  the  year  1872  Lie  has  put  forth  a 
series  of  memoirs  upon  the  most  difficult  and  most  advanced  parts  of  the 
integral  calculus.  He  commences  by  a  profound  study  of  the  works  of 
Jacobi  on  the  partial  differential  equations  of  the  first  order  and  at  first 
cooperates  with  Mayer  in  perfecting  this  theory  in  an  essential  point. 
Then,  by  continuing  the  study  of  this  beautiful  subject,  he  is  led  to 
construct  progressively  that  masterful  theory  of  continuous  transforma- 
tion groups  which  constitutes  his  most  important  work  and  in  which,  at 
least  at  the  start,  he  was  aided  by  no  one.  The  detailed  analysis  of 
this  vast  theory  would  require  too  much  space  here.  It  is  proper,  how- 
ever, to  point  out  particularly  two  elements  wholly  essential  to  these 
researches:  first,  the  use  of  contact  transformations  which  throws  such 

-'From  the  Bulletin  of  the  American  Mathematical  Society.    Translated  by  Edgar 
Odell  I<ovett  from  Comptcs  Rendus. 


452  FINKBIv'S   SOLUTION   BOOK. 

a  vivid  and  unexpected  light  upon  the  most  difficult  and  obscure  parts 
of  the  theories  relative  to  the  integration  of  partial  differential  equations; 
second,  the  use  of  infinitesimal  transformations.  The  introduction  of 
these  transformations  is  due  entirely  to  Lie;  their  use,  like  that  of 
Lagrange's  variation,  naturally  greatly  extends  both  the  notion  of  differ- 
ential and  the  applications  of  the  infinitesimal  calculus. 

The  construction  of  so  extended  a  theory  did  not  satisfy  Lie's  ac- 
tivity. In  order  to  show  its  importance  he  has  applied  it  to  a  great  num- 
ber of  particular  subjects,  and  each  time  he  has  had  the  good  fortune  of 
meeting  with  new  and  elegant  properties.  I  find  my  preference  in  the 
researches  which  he  has  published  since  1876  on  minimal  surfaces.  The 
theory  of  these  surfaces,  the  most  attractive  perhaps  that  presents  itself 
in  geometry,  still  awaits,  and  may  await  a  long  time,  the  complete  solu- 
tion of  the  first  problem  to  be  proposed  in  it,  namely,  the  determina- 
tion of  a  minimal  surface  passing  through  a  given  contour.  But,  in  re- 
turn, it  has  been  enriched  by  a  great  number  of  interesting  propositions 
due  to  a  multitude  of  geometers.  In  1866  Weierstrass  made  known  a 
very  precise  and  simple  system  of  formulae  which  has  called  forth  a 
whole  series  of  new  studies  on  these  surfaces.  In  his  works  Lie  returns 
simply  to  the  formulae  of  Monge ;  he  gives  their  geometric  interpreta- 
tion and  shows  how  their  use  can  lead  to  the  most  satisfactory  theory 
of  minimal  surfaces.  He  makes  known  methods  which  permit  of  deter- 
mining all  algebraic  minimal  surfaces  of  given  class  and  order.  Finally, 
he  studies  the  following  problem:  to  determine  all  algebraic  minimal 
surfaces  inscribed  in  a  given  algebraical  developable  surface.  He  gives 
the  complete  solution  for  the  case  where  only  one  of  these  surfaces  in- 
scribed in  the  developable  is  known. 

Of  great  interest  also  are  the  researches  which  we  owe  to  him  on 
the  surfaces  of  constant  curvature,  in  the  study  of  which  he  makes  use 
of  a  theorem  of  Bianchi  on  geodesic  lines  and  circles,  likewise  those  on 
surfaces  of  translation,  on  the  surfaces  of  Weingarten,  on  the  equations 
of  the  second  order  having  two  independent  variables,  et  cetera.  I  should 
reproach  myself  for  forgetting,  even  in  so  rapid  a  resume,  the  appli- 
cations which  Lie  has  made  of  his  theory  of  groups  to  the  non-Euclidean 
geometry  and  to  the  profound  study  of  the  axioms  which  lie  at  the  basis 
of  our  geometric  knowledge. 

These  extensive  works  quickly  attracted  to  the  great  geometer  the 
attention  of  all  those  who  cultivate  science  or  are  interested  in  its  prog- 
ress. In  1877  a  new  chair  of  mathematics  was  created  for  him  at  the 
University  of  Christiania,  and  the  foundation  of  a  Norwegian  review 
enabled  him  to  pursue  his  work  and  publish  it  in  full.  In  1886  he 
accepted  the  honor  of  a  call  to  the  University  of  Leipzig ;  he  taught  in 
this  university  with  the  rank  of  ordinary  professor  from  1886  to  1898. 
To  this  period  of  his  life  is  to  be  referred  the  publication  of  his  didactic 
works,  in  which  he  has  coordinated  all  his  researches.  Six  months  ago 
he  returned  to  his  native  land  to  assume  at  Christiania  the  chair  which 
had  been  especially  reserved  for  him  by  the  Norwegian  parliament,  with 
the  exceptional  salary  of  ten  thousand  crowns.  Unfortunately,  excess 
of  work  had  exhausted  his  strength  and  he  died  of  cerebal  anaemia  at  the 
age  of  fifty-six  years. 

Nowhere  is  his  loss  felt  more  keenly  than  in  our  country,  where  he 
had  so  many  friends.  True,  in  1870  a  misadventure  befell  him,  whose 
consequences  I  was  instrumental  in  averting.  Surprised  at  Paris  by  the 
declaration  of  war,  he  took  refuge  at'  Fontainebleau.  Occupied  in- 
cessantly by  the  ideas  fermenting  in  his  brain,  he  would  go  every  day 
into  the  forest,  loitering  in  places  most  remote  from  the  beaten  path, 
taking  notes  and  drawing  figures.  It  took  little  at  this  time  to  awaken 
suspicion.  Arrested  and  imprisoned  at  Fontainebleau,  under  conditions 
otherwise  very  comfortable,  he  called  for  the  aid  of  Chasles,  Bertrand, 
and  others;  I  made  the  trip  to  Fontainebleau  and  had  no  trouble  in 


BIOGRAPHY.  453 

convincing  the  procureur  imperial;  all  the  notes  which  had  been  seized 
and  in  which  figured  complexes,  orthogonal  systems,  and  names  of  ge- 
ometers, bore  in  no  way  upon  the  national  defenses.  Lie  was  released; 
his  high  and  generous  spirit  bore  no  grudge  against  our  country.  Not 
only  did  he  return  voluntarily  to  visit  it  but  he  received  with  great 
kindness  French  students,  scholars  of  our  Ecole  Normale  who  would 
go  to  Leipzig  to  follow  his  lectures.  It  is  to  the  Ecole  Normale  that 
he  dedicated  his  great  work  on  the  theory  of  transformation  groups.  A 
number  of  our  thesis  at  the  Sorbonne  have  been  inspired  by  his  teaching 
and  dedicated  to  him. 

The  admirable  works  of  Sophus  Lie  enjoy  the  distinction,  to-day 
quite  rare,  of  commanding  the  common  admiration  of  geometers  as  well 
as  analysts.  He  has  discovered  fundamental  propositions  which  will  pre- 
serve his  name  from  oblivion,  he  has  created  methods  and  theories  which, 
for  a  long  time  to  come,  will  exercise  their  fruitful  influence  on  the 
development  of  mathematics.  The  land  where  he  was  born  and  which 
has  known  how  to  honor  him  can  place  with  pride  the  name  of  Lie  be- 
side that  of  Abel,  of  whom  he  was  a  worthy  rival  and  whose  approaching 
centenary  he  would  have  been  so  happy  in  celebrating.  —  By  Professor 
Gaston  Darboux. 


454  FINKEI/S  SOLUTION  BOOK. 


BIOGRAPHY. 


SIMON  NEWCOMB,  PH.  DM  LL.  D. 

Simon  Newcomb  was  born  in  Wallace,  Nova  Scotia,  in  1835.  After 
being  educated  by  his  father  he  engaged  for  some  time  in  teaching.  He 
carne  to  the  United  States  in  1853,  and  was  engaged  for  two  years  as 
a  teacher  in  Maryland.  There  he  became  acquainted  with  Joseph  Henry 
and  Julius  E.  Hilgard,  who  recognizing  his  aptitude  for  mathematics, 
secured  his  appointment  in  1857  as  computer  on  the  "Nautical  Almanac," 
which  was  then  published  in  Cambridge,  Mass.  In  Cambridge  he  came 
under  the  influence  of  Professor  Benjamin  Peirce.  He  entered  the  Law- 
rence Scientific  School  and  was  graduated  in  1858,  continuing  thereafter 
for  three  years  as  a  graduate  student. 

In  1861  he  was  appointed  professor  of  mathematics  in  the  U.  S. 
Navy  and  assigned  to  duty  at  the  U.  S.  Naval  Observatory  in  Wash- 
ington. There  he  negotiated  the  contract  for  the  26-inch  equatorial  tele- 
scope authorized  by  congress,  supervised  its  construction  and  planned  the 
tower  and  dome  in  which  it  is  mounted. 

He  was  chief  director  of  the  commission  created  by  congress  to  ob- 
serve the  transit  of  Venus  on  December  8,  1874.  He  visited  the  Sas- 
katchewan region  in  1860  to  observe  an  eclipse  of  the  Sun,  and  in  1870-1 
was  sent  to  Gibralter  for  a  similar  purpose.  In  1882  he  commanded  an 
expedition  to  observe  the  transit  of  Venus  at  the  Cape  of  Good  Hope. 
Meanwhile  in  1887  he  became  senior  professor  of  mathematics  in  the  U.  S. 
Navy,  and  since  that  time  has  been  in  charge  of  the  office  of  the  "Amer- 
ican Ephemeris  and  Nautical  Almanac."  Professor  Newcomb  has  a  large 
corps  of  assistants  in  Washington. 

In  addition  to  these  duties,  in  1884  he  became  professor  of  mathe- 
matics and  astronomy  in  Johns  Hopkins,  (succeeding  the  distinguished 
Sylvester,  upon  the  departure  of  the  latter  to  accept  a  professorship  at 
Oxford),  where  he  has  had  charge  of  the  American  Journal  of  Mathe- 
matics. However  he  is  not  now  editor  of  that  Journal,  having  recently 
severed  his  immediate  active  connection  with  the  Johns  Hopkins  Uni- 
versity for  the  next  two  or  three  years. 

Professor  Newcomb  has  been  intimately  associated  with  the  equip- 
ment of  the  Lick  observatory  of  California,  and  examined  the  glass  of 
the  great  telescope  and  its  mounting  before  its  acceptance  by-  the  trustees. 

The  results  of  his  scientific  work  have  been  given  to  the  world  in 
more  than  one  hundred  papers  and  memoirs.  Concerning  these,  Arthur 
Cayley,  president  of  the  Royal  Astronomical  Society  of  Great  Britain,  said: 
"Professor  Newcomb's  writings  exhibit,  all  of  them,  a  combination  on 
the  one  hand  of  mathematical  skill  and  power  and  on  the  other  of  good 
hard  work,  devoted  to  the  furtherance  of  astronomical  science." 

His  work  has  been  principally  in  the  mathematical  astronomy  of  the 
solar  system,  particularly  Neptune,  Uranus,  and  the  Moon,  but  the  whole 
plan  includes  the  most  exact  possible  tables  of  the  motions  of  all  the 
planets.  Amongst  the  most  important  of  his  papers  are :  "On  the  Secu- 
lar Variations  and  Mutual  Relations  of  the  Orbits  of  the  Asteroids" 
(1860)  ;  "An  Investigation  of  the  Orbit  of  Neptune,  with  general  tables 
of  its  motion"  (1874)  ;  "Researches  on  the  Motion  of  the  Moon"  (1876)  ; 
"Measure  of  the  Velocity  of  Light"  (1884)  ;  and  "Development  of  the 
Purturbative  Function  and  its  Derivative  in  the  Sines  and  Cosines  of  the 
Eccentric  Anomaly,  and  in  Powers  of  the  Eccentricities  and  Inclinations" 
(1884). 

In  1874  Columbian  University  of  Washington  conferred  on  him 'the 
degree  of  LL.  D.,  and  in  1875  he  received  the  same  degree  from  Yale,, 


BIOGRAPHY.  455 

also  from  Harvard  in  1874,  and  from  Columbia  College  in  1887,  while  on 
the  300th  anniversary  of  the  founding  of  the  University  of  Leyden  in 
1875,  that  institution  gave  him  him  the  degree  of  Master  of  Mathematics 
and  Doctor  of  Natural  Philosophy,  and  on  the  500th  anniversary  of  the 
University  of  Heidelberg  in  1886  he  received  the  degree  of  Ph.  D.  Be- 
sides the  degrees  just  mentioned  he  received  one  from  Edinburgh  in  1891, 
one  on  the  occasion  of  the  tercentenary  of  the  University  of  Dublin  in 
1892,  and  one  from  Paris  on  the  tercentenary  of  Galileo's  connection  with 
the  University  in  1893. 

He  was  awarded  the  gold  medal  of  the  Royal  Astronomical  Society  in 
1874  and  in  1878  received  the  great  gold  Huyghens  medal  of  the  Uni- 
versity of  Leyden,  which  is  given  to  astronomers  once  in  20  years  for 
the  most  important  work  accomplished  in  that  science  between  its 
awards.  Besides  the  two  gold  medals  mentioned  Professor  Newcomb 
received  a  third  in  1890,  the  Copley  medal,  given  by  the  Royal  Society 
of  England. 

In  1887  the  Russian  Government  ordered  the  portrait  of  Professor 
Newcomb  to  be  painted  for  the  collection  of  famous  astronomers  at  the 
Russian  observatory  at  Pulkowa,  and  also  ordered  to  be  presented  to 
him  a  vase  of  jasper  with  marble  pedestal  seven  feet  high.  The  Univer- 
sity of  Tokyo  has  also  presented  him  with  two  vases  of  bronze. 

He  was  elected  an  associate  member  of  the  Royal  Astronomical  So- 
ciety in  1872,  corresponding  member  of  the  Institute  of  France  in  1874, 
and  foreign  member  of  the  Royal  Society  1877 ;  and  he  also  holds  hon- 
orary or  corresponding  relations  to  nearly  all  the  European  academies  of 
Science.  In  1877  he  was  elected  one  of  the  eight  members  of  the  council 
of  the  Astronomische  Gesellschaft,  an  international  astronomical  society 
that  meets  once  in  two  years.  He  was  elected  to  the  National  Academy 
of  Sciences  in  1869  and  since  1883  has  been  its  vice  president.  In  1876 
he  was  elected  president  of  the  American  Association  for  the  Advance- 
ment of  Science,  and  delivered  his  retiring  address  at  the  St.  Louis 
meeting  in  1878.  He  also  held  the  presidency  of  the  American  Society 
for  Physical  Research. 

He  was  elected  member  of  the  New  York  Mathematical  Society  in 
1891,  and  delivered  an  address,  entitled  "Modern  Mathematical  Thought" 
before  the  annual  meeting  of  the  Society,  December  28,  1893,  which  was 
published  in  the  Bulletin  of  the  Society  for  January  1894,  and  in  Nature 
of  February  1,  1894. 

Professor  Newcomb's  book  on  Popular  Astronomy  (1877)  has  been 
republished  in  England  and  translated  into  German,  while  "School  As- 
tronomy" by  Newcomb  and  Holden  (1879),  and  their  "Briefer  Course" 
(1883),  are  used  as  text  books  in  most  of  our  colleges. 

Professor  Newcomb  has  also  carried  on  important  investigations  on 
subjects  purely  mathematical.  An  important  contribution  by  him  on 
"Elementary  Theorems  Relating  to  the.  Geometry  of  a  Space  of  Three 
Dimensions  and  of  Uniform  Positive  Curvature  in  the  Fourth  Dimen- 
sion," was  published  in  Borchardt's  Journal,  Berlin,  1877.  Full  extracts 
of  this  important  contribution  to  non-Euclidean  geometry  are  given  in 
the  Encyclopaedia  Britannica,  article  "Measurment."  In  Vol.  I.  of  the 
American  Journal  of  Mathematics  he  has  a  note  "On  a  Class  of  Trans- 
formations which  Surfaces  may  Undergo  in  Space  of  more  than  Three 
Dimensions,"  in  which  he  shows,  for  instance,  that  if  a  fourth  dimen- 
sion were  added  to  space,  a  closed  material  surface  (or  shell)  could  be 
turned  inside  out  by  simple  flexure  without  either  stretching  or  tearing. 
Later  articles  have  been  on  the  theory  of  errors  in  observations.  In 
former  years  he  also  contributed  to  the  Mathematical  Monthly  and  the 
Analyst. 

He  has  also  written  a  series  of  mathematical  text-books,  comprising 
Algebra  (1881)  ;  Geometry  (1881)  ;  Trigonometry  and  Logarithms 


456  FINKEI/S  SOLUTION  BOOK. 

(1882) ;  School  Algebra  (1882)  ;  Analytic  Geometry  (1884)  ;  Essentials 
of  Trigonometry  (1884);  and  Calculus  (1887).  These  works  have  been 
favorably  received  and  are  everywhere  regarded  as  text-books  of  decided 
merits. 

Professor  Newcomb  refers  to  astronomy  as  his  profession  and  to  polit- 
ical economy  as  his  recreation,  and  in  the  latter  branch  has  written  several 
books  and  a  number  of  magazine  articles.  —  By  J.  M.  Colaw.  From  the 
American  Mathematical  Monthly,  Vol.  I,  No.  8. 


BIOGRAPHY. 


457 


GEORGE  BRUCE  HALSTED. 

Dr.  Halsted,  a  direct  descendant  of  Abram  Clark,  signer  of  the 
declaration  of  independence,  was  born  in  Newark,  N.  J.,  November  25, 
1853.  The  Halsteds  have  been  for  four  generations  graduates  of  Prince- 
ton. Winning  the  Mathematical  Fellowship  at  Princeton,  Halsted  went 
for  applied  mathematics  to  Columbia  School  of  Mines,  won  while  there 
an  intercollegiate  prize,  and  was  made  one  of  the  first  Fellows  of  the 
new  Johns  Hopkins  University.  Sylvester  was  particularly  partial  to 
him,  and  Halsted's  life  of  Sylvester  is  recognized  as  the  authority  on 
the  American  part  of  his  career.  Says  Dr.  Fabian  Franklin:  "Professor 
Halsted,  in  his  account  of  Sylvester's  work  already  referred  to,  points  out 
how  the  vicissitudes  of  his  career  were  reflected  in  the  richness  or  the 
meagreness  of  his  mathematical  production  from  period  to  period." 
Says  Major  P.  A.  MacMahon  in  the  biography  of  Sylvester  published 
by  the  Royal  Society:  "Sylvester's  first  high  class  consisted  of  but  one 
student,  G.  B.  Halsted.  This  gentleman,  since  well  known  in  science, 
had  the  most  beneficial  effect  upon  his  master,  for  it  was  owing  to  his 
enthusiasm  and  persistence  that  Sylvester's  attention  was  again  called 
to  the  Modern  Higher  Algebra  and  the  Theory  of  Invariants,  and  a  fruit- 
ful crop  of  new  discoveries  was  almost  the  immediate  result.  Before 
taking  his  Doctor's  Degree  at  the  Johns  Hopkins  University,  Halsted 
published  in  1878  his  epoch-making  Bibliography  of  Hyper-Space  and 
Non-Euclidean  Geometry,  long  out  of  print  and  greatly  in  demand,  which 
began  at  once  to  be  cited  all  over  the  world  and  was  reproduced  in 
Russia.  Called  to  introduce  modern  higher  mathematics  at  his  ancestral 
college,  Princeton,  the  papers  set  on  quaternions,  determinants,  modern 
higher  algebra,  and  history  of  mathematics  by  Dr.  Halsted  were  the 
first  ever  given  at  Princeton.  His  teaching  of  the  History  of  Mathe- 
matics attracted  attention  as  early  as  1881,  when  he  had  in  his  class  Pro- 
fessor H.  B.  Fine  and  Professor  A.  L.  Kimball.  His  work  in  history  of 
mathematics  has  never  since  been  discontinued.  While  at  Princeton,  Dr. 
Halsted  produced  his  treatise  on  Mensuration  which  had  the  honor  of 
being  drawn  upon  by  Professor  Wm.  Thomson  for  his  article  "Mensura- 
tion" in  the  ninth  edition  of  the  Encyclopaedia  Britannica.  He  took 
from  it,  among  other  novelties,  the  steregon,  the  steradian,  and  the  treat- 
ment of  solid  angles  associated  with  them.  Simon  Newcomb  wrote  of 
Dr.  Halsted:  "He  is  the  author  of  a  treatise  on  Mensuration  which  is 
the  most  thorough  and  scientific  with  which  I  am  acquainted." 

From  this  brilliant  field  of  work  Dr.  Halsted  was  called  by  a  tele- 
gram from  the  University  of  Texas,  announcing  his  election  to  the  pro- 
fessorship of  mathematics,  urging  his  acceptance.  Dr.  Halsted  had  al- 
ready produced  17  scientific  papers  on  Mathematics  and  Logic.  This 
productivity  was  not  abated  by  his  removal  to  Texas,  and  the  titles  of  his 
papers  now  approach  a  hundred. 

His  Elements  of  Geometry  which  appeared  in  1885,  has  passed 
through  seven  editions.  The  section  headed  by  his  phrase  "Partition  of 
a  Perigon,"  has  become  classic. 

Of  Dr.  Halsted's  Elementary  Synthetic  Geometry,  which  appeared 
in  1893,  has  been  said:  "For  more  than  two  thousand  years  geometry 
has  been  founded  upon,  and  built  up  by  means  of,  congruent  triangles. 
At  last,  after  twenty  centuries,  comes  a  book  reaching  all  the  preceding 
results  without  making  any  use  of  congruent  triangles;  and  so  simply 
that,  for  example,  all  the  ordinary  cases  of  congruence  of  triangles  are 
demonstrated  together  in  eight  lines." 


458  FINKEIv'S  SOLUTION  BOOK. 

Dr.  Halsted's  two  books  are  as  yet  the  only  geometries  treating 
spherics  comparatively  by  considering  the  sphere  as  the  analogue  of 
the  plane.  "In  the  method  employed  by  Dr.  Halsted  almost  the  whole  of 
the  geometry  of  the  plane  is  directly  applicable  to  the  sphere."  The  Ele- 
mentary Synthetic  Geometry  contains  an  introduction  to  the  new  Le- 
moine-Brocard  Geometry,  the  only  one  which  has  appeared  on  the 
western  continent. 

David  Eugene  Smith,  in  his  history  of  Modern  Mathematics  men- 
tions, p.  567,  as  "among  the  most -active"  in  securing  the  acceptance  of 
the  Bolyai-LobachevskHdea,  Houel  in  France,  Riemann,  Helmholtz  and 
Baltzer  in  Germany,  de  Tilley  (1879)  in  Belgium,  Clifford  in  England, 
and  Halsted  (1878)  in  America. 

Of  these  all  are  dead  but  Halsted.  By  Halsted  alone  were  the  im- 
mortal works  of  Saccheri,  Bolyai,  Lobachevski,  Vasiliev,  made  available 
to  the  English  speaking  world,  and  also  Japan,  for  his  Bolyai  and  Lo- 
bachevski have  been  reissued  in  Tokio.  Results  of  his  travels  in  Japan, 
Hungary,  Russia,  added  fuel  to  the  awakening  fire  of  interest  in  all  things 
non-Euclidean,  and  now  the  majestic  face  of  heroic  Lobachevski  has 
t>een  worthily  given  to  the  world  in  the  beautiful  picture  made  from  a 
photograph  furnished  by  Dr.  Halsted  to  the  Open  Court  as  frontispiece 
for  his  Life  of  Lobachevski  in  the  number  for  July  1898. 

Of  John  Bolyai  he  could  find  in  all  Hungary  no  picture;  his  linea- 
ments are  lost  forever. 


BIOGRAPHY. 
BIOGRAPHY, 


PROFESSOR  FELIX  KLEIN. 

The  eminent  subject  of  this  very  imperfect  sketch  was  born  on  the 
.twenty-fifth  of  April,  1849,  in  Duesseldorf.  His  mother  was  Elise  Sophie 
nee  Kayser;  his  father,  the  "Landrentmeister"  Caspar  Klein,  both  of 
the  protestant  faith.  For  eight  years,  from  the  autumn  of  1857  to  the 
autumn  of  1865  he  attended  the  Duesseldorf  Gymnasium,  and  went  thence 
to  the  University  of  Bonn,  for  the  study  of  mathematics  and  the  natural 
sciences,  especially  physics.  Here  he  had  the  extraordinary  good  fortune 
to  come  into  close  relations  with  the  great  Professor  Pluecker,  who  gave 
him  the  position  of  assistant  in  the  physical  institute  of  Bonn,  and  used 
his  help  in  writing  out  his  profoundly  original  and  stimulating  mathe- 
matical works. 

The  death  of  Pluecker  May  22nd,  1868,  closed  this  formative  period, 
of  which  the  influence  on  Klein  can  not  be  overestimated.  So  mighty  is 
the  power  of  contact  with  the  living  spirit  of  research,  of  taking  part  in 
original  work  with  a  master,  of  sharing  in  creative  authorship,  that  any 
one  who  has  once  come  intimately  in  contact  with  a  producer  of  the 
first  rank  must  have  had  his  whole  mentality  altered  for  the  rest  of 
his  life. 

The  gradual  development,  high  attainment,  and  then  continuous 
achievement  of  Felix  Klein  are  more  due  to  Pluecker  than  to  all  other 
influences  combined.  His  very  mental  attitude  in  the  world  of  mathe- 
matics constantly  recalls  his  great  maker. 

Of  others  whose  lectures  he  attended,  we  may  mention  Argelander 
<and  Lipschitz,  to  the  latter  of  whom  particularly  he  has  expressed  his 
gratitude  for  kindly  and  efficient  guidance  and  aid  in  his  studies.  Klein 
took  his  doctor's  degree  at  Bonn  on  December  12th,  1868,  with  a  disser- 
tation "On  the  transformation  of  the  general  equation  of  the  second 
degree  between  line-coordinates  to  a  canonic  form,"  a  subject  taken  from 
the  analytic  line-geometry  of  his  master  Pluecker.  A  line-complex  of  the 
nth  degree  contains  a  triply  infinite  multitude  of  straights,  which  are 
so  distributed  in  space,  that  those  straights  which  go  through  a  fixed 
point  make  a  cone  of  the  nth  order,  or,  what  is  the  same,  that  those 
straights  which  lie  in  a  fixed  plane  envelop  a  curve  of  the  nth  class.  Such 
an  aggregate  or  form  finds  its  analytic  representation  through  the  coordi- 
nates of  the  straight  in  space,  introduced  by  Pluecker.  According  to 
Pluecker  the  straight  has  six  homogeneous  coordinates  which  fulfill  an 
equation-pf-conditipn  of  the  second  degree.  By  means  of  these  the 
straight  is  determined  with  reference  to  a  coordinate-tetrahedron.  A 
homogeneous  equation  of  the  nth  degree  between  these  coordinates  rep- 
resents a  complex  of  the  nth  degree. 

The  dissertation  transforms  the  equation  of  the  second  degree  be- 
tween line-coordinates  to  a  canonic  form,  in  correspondence  with  a 
change  of  the  coordinate-tetrahedron.  It  first  gives  the  general  formulas 
to  be  applied  in  such  a  transformation. 

From  these  the  problem  appears  algebraically  as  the  simultaneous 
linear  transformation  of  the  complex  to  a  canonic  form,  and  of  theequation- 
'pf-condition,  which  the  line  coordinates  must  fulfill,  into  itself.  In  carry- 
ing out  these  transformations,  it  attains  to  a  classification  of  the  com- 
plexes of  the  second  degree  into  distinct  species. 

The  dissertation  is  dedicated  to  Pluecker  and  contains  eight  specific 
references  to  Pluecker's  "Neue  Geometric  des  Raumes,  gegruendet  auf  die 
Betrachtung  der  geraden  Linie  als  Raumelement."  It  is  lucid  and  simple, 
but  for  depth  and  promise  contrasts  sharply  with  the  great  dissertation 
•of  Riemann,  that  "book  with  seven  seals." 


460  FINKEI/S  SOLUTION  BOOK. 

It  may  be  interesting,  as  characteristic  of  this  germinating  state,  to 
note  that  of  his  five  theses  the  second  calls  attention  to  one  of  Cauchy's 
slips  in  logical  rigor,  slips  now  known  to  be  so  numerous  that  C.  S.  Pierce 
makes  of  them  a  paradox,  maintaining  that  fruitfulness  of  Cauchy's  work 
is  essentially  connected  with  its  logical  inaccuracy. 

The  third  thesis  declares  the  assumption  of  an  ether  unavoidable  in 
the  explanation  of  the  phenomena  of  light. 

The  last  thesis  is  the  desirability  of  the  introduction  of  newer  methods 
in  Geometry  alongside  the  Euclidean  in  gymnasial  teaching. 

This  serves,  it  seems,  to  emphasize  my  point  that  the  long  eight  years 
of  gymnasial  so-called  training  left  the  seed  still  dormant,  and  only  in 
Pluecker  did  it  find  the  rain  and  the  sun  to  call  it  to  life  and  growth. 

Within  two  years  now  the  development  is  amazing.  Already  in  1870 
he  is  working  with  another  great  genius,  Sophus  Lie ;  and  in  1871  is 
presented  to  the  Goettingen  Academy  of  Science  his  epoch-making  paper, 
"Ueber  die  sogenannte  Nicht-Euklidische  Geometric."  Its  aim  is  to  pre- 
sent the  mathematical  results  of  the  non-Euclidean  geometry,  in  so  far 
as  they  pertain  to  the  theory  of  parallels,  in  a  new,  intuitive  way ;  its  in- 
strument is  the  mighty  projective  geometry,  which  he  proves  independent 
of  all  question  of  parallels.  He  perfects  the  projective  metrics  of  Cay- 
ley  by  founding  cross-  ratio,  after  von  Staudt,  wholly  without  any  use  or 
idea  of  measurement.  Then  can  be  constructed  a  general  projective  ex- 
pression for  distance,  related  to  an  arbitrary  surface  of  the  second  degree 
as  Fundamental-surface  (Cayley's  Absolute).  This  projective  metrics 
then  gives,  according  to  the  species  of  Absolute  used,  a  picture  of  the 
results  of  the  parallel-theory  in  the  space  of  Lobachevsky,  of  Euclid,  of 
Riemann.  But  not  merely  a  picture ;  they  coincide  to  their  innermost 
nature. 

The  paper  begins  by  stating  that,  as  well-known,  the  eleventh  axiom 
of  Euclid  is  equivalent  to  the  theorem  that  the  sum  of  the  angles  in  a 
triangle  equals  two  right  angles.  Legendre  gave  a  proof  that  the  angle- 
sum  in  a  triangle  cannot  be  greater  than  two  right  angles ;  but  this  proof, 
like  the  corresponding  one  in  Lobachevsky,  assumes  the  infinite  length  of 
the  straight. 

Drop  this  assumption,  and  the  proof  falls,  else  would  it  apply  in 
surface  spherics.  Legendre  showed  further,  that  if  in  one  triangle  the 
angle-sum  is  two  right  angles,  it  is  so  in  every  triangle.  We  now  know 
that  this  had  been  proven  long  before  by  Saccheri.  But  Professor  Klein 
said  that  he  heard  the  name  of  Saccheri  for  the  first  time  in  my  address 
before  the  World's  Science  Congress.  But  it  is  claimed  for  Gauss  that  he 
was  the  first  to  distinctly  state  his  conviction  of  the  impossibility  of  prov- 
ing the  theorem  of  the  equality  of  the  angle-sum  to  two  right"  angles. 
But  it  does  not  follow,  as  claimed  by  his  Goettingen  worshippers,  that 
Gauss  ever  came  to  the  ^ conviction  that  a  valid  non-Euclidean  geometry 
was  possible  until  after  it  had  been  made  simultaneously  by  John  Bolyai 
and  Lobachevsky,  and  perhaps  long  before  by  Wolfgang  Bolyai.  Cer- 
tainly the  world  did  not  hear  of  it  from  Gauss.  He  published  nothing 
on  it. 

In  this  non-Euclidean  geometry  there  appears  a  certain  constant 
characteristic  for  the  metrics  of  the  space.  By  giving  this  an  infinite 
value  we  obtain  the  ordinary  Euclidean  geometry.  But  if  it  has  a  finite 
value,  we  get  a  quite  distinct  geometry,  in  which,  for  example,  the  follow- 
ing theorems  hold:  The  angle-sum  in  a  triangle  is  less  than  two  right 
angles,  and  indeed  so  much  the  more  so  the  greater  the  surface  of  the 
triangle.  For  a  triangle  whose  vertices  are  infinitelv  separated,  the  angle- 
sum  is  zero.  Through  a  point  without  a  straight  one  can  draw  two 
parallels  to  the  straight,  that  is,  lines  which  cut  the  straight  on  the  one 
or  the  other  side  in  a  point  at  infinity.  The  straights  through  the  point 
which  run  between  the  two  parallels  nowhere  cut  the  given  straight.  But 
on  the  other  hand,  in  Riemann' s  marvellous  inaugural  lecture,  "Ueber  die- 


BIOGRAPHY. 


461 


Hypothesen,  welche  der  Geometric  zu  Grunde  liegen,"  is  pointed  out  that 
the  unboundedness  of  space,  which  is  experiential,  does  not  carry  with  it 
the  infinity  of  space. 

It  is  thinkable,  and  would  not  contradict  our  perceptional  intuition, 
which  always  relates  to  a  finite  piece  of  space,  that  space  is  finite  and 
comes  back  into  itself. 

The  geometry  of  our  space  would  then  be  like  that  of  a  tridimensioual 
sphere  in  a  four  dimensional  manifoldness.  This  representation  carries 
with  it  that  the  angle-sum  in  a  triangle,  as  in  ordinary  spherical  triangles, 
is  greater  than  two  right  angles,  and  indeed  the  more  so,  the  greater  the 
triangle.  The  straight  would  then  have  no  point  at  infinity,  and  through 
a  given  point  no  parallel  to  a  given  straight  could  be  drawn.  Now  Cayley 
constructed  his  celebrated  projective  metrics  to  show  how  the  ordinary 
Euclidean  metrics  may  be  taken  as  a  special  part  of  projective  geometry. 
Klein  generalizes  Cayley  and  founds  three  metric  geometries,  the  elliptic 
(Riemann's),  the  hyperbolic  (Lobachevsky's),  the  parabolic  (Euclid's). 

This  little  paper  of  1871  contains  the  promise  of  much  that  is  most 
genial  in  the  after  work  of  a  man  now  generally  considered  as  the  most 
interesting  and  one  of  the  very  greatest  of  living  mathematicians.  Of  all 
those  splendid  and  charming  series  of  lectures  with  which  Klein  has  made 
Goettingen  so  attractive  to  the  whole  world,  the  most  delightful  and 
epoch-making  are  those  on  non-Euclidean  geometry,  (Nicht-Euklidische 
Geometric,  I.  Vorlesung,  gehalten  waehrend  des  Wintersemesters  1889-90 
von  F.  Klein.  Ausgearbeitet  von  Fr.  Schilling.  Zweiter  Abdruck.  Goet- 
tingen, 1893.  Small  Quarto,  lithographed,  pp.  v.  365.  II.  Sommersemes- 
ters  1890.  Zweiter  Abdruck  1893,  pp.  iv.  238). 

The  World's  Science  Congress  at  Chicago  was  in  nothing  more  fortu- 
nate than  in  the  presence  of  Helmholtz  and  Felix  Klein,  and  in  the  spon- 
taneous and  universal  homage  accorded  them  no  idea  was  more  often 
emphasized  than  their  connection  with  the  birth  and  development  of  that 
wonderful  new  world  of  pure  science  typified  in  the  non-Euclidean 
geometry. 

The  narrow  limits  of  this  feeble  sketch  prevent  the  statement  of  how 
much  promise,  richly  fulfilled  in  the  development  of  this  many-sided  man, 
in  totally  other  directions  is  contained  in  a  little-known  paper  of  1873, 
"Ueber  den  allgemeinen  Functionsbegriff  und  dessen  Darstellung  durch 
eine  willkuerliche  Curve." 

Twenty  years  of  production  and  achievement  have  not  in  the  least 
dampened  the  ardour  of  this  enthusiastic  mind.  This  very  summer  at  the 
great  meeting  of  scientists  in  Vienna  Klein  seemed  the  busiest,  the  fore- 
most of  all  that  goodly  company.  —  By  Dr.  George  Bruce  Halsted.  From 
the  American  Mathematical  Monthly,  Vol.  I,  No.  12. 


462  FINKEL'S  SOLUTION  BOOK. 

BIOGRAPHY. 


BENJAMIN  PEIRCE. 

Benjamin  Peirce  was  born  at  Salem,  Massachusetts,  April  4,  1809, 
and  died  at  Cambridge,  Massachusetts,  October  6,  1880.  He  entered  Har-. 
vard  College,  at  the  age  of  sixteen;  and,  at  the  age  of  twenty,  he  was 
graduated  from  the  same  College,  with  highest  honors.  He  devoted  him- 
self principally  to  the  study  of  Mathematics.  This  favorite  study  of  hia 
was  pursued  far  beyond  the  limits  of  the  curriculum  of  mathematical 
studies  prescribed  by  the  authorities  of  Harvard  College,  at  that  time. 

As  an  under-graduate  student,  young  Peirce  was  instructed  by 
Nathaniel  Bowditch,  who  soon  perceived  the  innate  mathematical  genius 
of  his  pupil.  Bowditch  proudly  predicted  the  future  greatness  of  the 
young  man.  Not  only  did  Bowditch  give  him  valuable  instruction  in 
geometry  and  analytics,  but  also  acted  as  his  mathematical  adviser  — 
carefully  directing  him  in  the  development  of  his  mathematical  talents  and 
scientific  powers.  The  lectures  on  higher  mathematics  delivered  by  Francis 
Grund  he  was  enabled  to  attend,  by  reason  of  his  preparation  beyond  the 
limit  of  the  under-graduate  course  in  mathematics.  When  Dr.  Bowditch 
was  publishing  his  translation  and  commentary  of  the  Mehanique  Celeste 
of  Laplace,  young  Peirce  assisted  in  reading  the  proof-sheets.  This  crit- 
ical reading  of  that  great  work  of  Laplace  was  to  him  an  education  in 
itself,  and  may  have  been  the  prime  cause  that  not  a  small  part  of  Peirce's 
subsequent  mathematical  and  scientific  work  was  done  in  the  great  field 
of  analytical  mechanics : 

In  the  class-room,  he  frequently  gave  original  demonstrations  which 
proved  to  be  more  direct  and  scientific  than  those  given  in  the  text-books 
of  that  day.  On  graduating,  he  went  to  Northampton,  Massachusetts,  as 
a  teacher  in  Mr.  Bancroft's  School.  As  tutor,  he  returned  to  Harvard 
College,  in  1831.  Since  Professor  Farrar  spent  the  next  year  in  Europe, 
tutor  Peirce  was  left  at  the  head  of  the  Department  of  Mathematics  in 
Harvard  College;  and,  on  account  of  the  physical  inability  of  Professor 
Farrar  to  resume  teaching,  Peirce  continued  to  fill  his  place.  In  fact, 
Peirce  held  this  position,  advancing  step  by  step,  until  the  time  of  his 
death.  His  position,  in  1842,  was  christened  "The  Perkins  Professorship 
of  Mathematics  and  Astronomy."  In  the  history  of  mathematical  teach- 
ing at  Harvard  College,  the  year  1833  marks  an  important  epoch;  as  it 
was  then  that  Benjamin  Peirce  became  the  professor  of  Mathematics  and 
Natural  Philosophy  in  that  institution  of  learning. 

Professor  Peirce  was  married  in  July,  1833.  At  the  time  of  his  death, 
there  were  living  his  wife,  three  'sons,  and  a  daughter.  His  eldest  son, 
James  M.  Peirce,  is  University  professor  of  mathematics  in  Harvard; 
Charles  S.  Peirce  is  a  professor  in  the  Johns  Hopkins  University;  and 
H.  H.  D.  Peirce  is  connected  with  the  firm  of  Herter  Brothers,  New 
York  City. 

It  has  been  said  that  a  mere  boy  detected  an  error  in  Bowditch's  solu- 
tion of  a  problem.  "Bring  me  the  boy  who  corrects  my  mathematics," 
said  Bowditch.  Master  Benjamin  Peirce  was  the  boy  who  had  done  the 
correcting;  and  thirty  years  later,  this  same  Benjamin  Peirce  dedicated 
one  of  his  great  mathematical  works  "To  the  cherished  and  revered 
memory  of  my  master  in  science,  Nathaniel  Bowditch,  The  Father  of 
American  Geometry."  This  same  title  was  bestowed  upon  Peirce,  by  for- 
eign mathematicians.  Sir.  "Win.  Thomson  (Lord  Kelvin),  in  an  address 
before  the  British  Association,  referred  to  Benjamin  Peirce  as  "The 
Founder  of  High  Mathematics  in  America;"  and  on  a  similar  occasion, 
the  late  Professor  Cayley  referred  to  him  as  "The  Father  of  American 
Mathematics."  The  name  of  Benjamin  Peirce  is  that  of  an  American 


BIOGRAPHY.  463 

mathematician,  whom  no  one  need  hesitate  to  rank  with  the  names  of 
Pythagoras,  Leibnitz,  Newton,  Legendre,  John  Bernoulli,  Wallis,  Abel, 
Laplace,  Lagrange,  and  Euler.  Through  the  united  efforts  of  the  late 
Professor  Wm.  Chauvenet  (Yale's  ablest  mathematician  and  astronomer) 
and  Benjamin  Peirce  —  not  to  speak  of  their  worthy  successors,  was 
effected  the  general  adoption  of  the  ratio-system  in  American  works  on 
trigonometry. 

In  the  reforms  incident  to  the  New  Education,  Harvard  has  always 
taken  a  prominent  part  and  Benjamin  Peirce  was  an  enthusiastic  advocate 
of  the  elective  system  with  respect  to  collegiate  studies.  As  a  branch  of 
Harvard  College,  there  was  opened,  in  1842,  the  Lawrence  Scientific 
School ;  and  in  this  school,  Professor  Peirce  gave  instruction  in  higher 
Mathematics  including  analytical  and  celestial  mechanics.  Such  advanced 
courses  of  mathematics,  as  he  offered  to  students,  in  1848,  had  never 
before  been  offered  to  American  students  by  any  other  professor  in  any 
other  American  college.  The  second  American  educational  institution 
which  offered  equally  advanced  courses  of  mathematics,  is  the  Johns 
Hopkins  University;  and  these  courses  were  arranged  by  that  English 
master,  who  gave  a  fresh  and  powerful  impulse  to  mathematical  study 
and  teaching  in  America  —  Professor  J.  J.  Sylvester. 

The  preparation  of  mathematical  text-books  was  begun  by  Professor 
Peirce,  immediately  on  beginning  his  career  as  teacher  of  Mathematics  in 
Harvard  College.  In  1835  appeared  his  Elementary  Treatise  on  Plane 
Trigonometry;  in  1836,  his  Elementary  Treatise  on  Spherical  Trigo- 
nometry together  with  his  Elementary  Treatise  on  Sound;  in  1837,  his 
Elementary  Treatise  on  Plane  and  Solid  Geometry  together  with  his  Ele- 
mentary Treatise  on  Algebra;  during  the  period  of  1841^6,  he  wrote 
and  published  in  two  volumes  his  Elementary  Treatise  on  Curves,  Func- 
tions, and  Forces;  and  in  1855,  he  published  his  Analytical  Mechanics. 
Subsequently  was  published  his  memoir  on  Linear  Associative  Algebra; 
and  this  memoir,  according  to  Professor  James  Mills  Peirce,  he  regarded 
as  his  great  work.  All  of  his  works  are  models  of  conciseness,  perspicuity, 
and  elegance;  and  they  all  evince  extraordinary  originality  and  genius. 

In  1867,  Professor  Peirce  was  made  the  Superintendent  of  the  United 
States  Coast  Survey;  and  he  held  that  position  for  seven  years.  He  had 
been  consulting  astronomer  to  the  American  Ephemeris  and  Nautical  Al- 
manac, since  1849;  and  for  many  years,  he  directed  the  theoretical  part 
of  the  work.  In  1855,  Professor  Peirce  was  one  of  the  men  intrusted 
with  the  organization  of  the  Dudley  Observatory.  For  many  years  before 
and  after  he  took  charge  of  the  United  States  Coast  Survey,  he  was 
frequently  consulted  with  respect  to  the  work  in  that  office.  He  received 
the  degree  of  Doctor  of  Laws  from  the  University  of  North  Carolina,  in 
1847,  and  also  from  Harvard  University  in  1867.  He  was  elected  an  As- 
sociate of  the  Royal  Astronomical  Society  of  London  in  1849,  and  a 
member  of  the  Royal  Society  of  London  in  1852.  He  was  elected  presi- 
dent of  the  American  Association  for  the  Advancement  of  Science,  in 
1853  (the  fifth  year  of  its  existence)  ;  and  he  was  one  of  the  original 
members  of  the  Royal  Societies  of  Edinburg,  and  Goettingen;  Honorary 
Fellow  of  the  Imperial  University  of  St.  Vladimir,  at  Kiev ; "  etc. 

Professor  Peirce's  conception  of  the  American  Social  Science  Asso- 
ciation was  that  it  should  be  a  university  for  the  people,  —  combining  those 
who  can  contribute  any  thing  original  in  social  science  into  a  temporary 
academical  senate,  to  meet  for  some  weeks  in  a  given  place  and  debate 
questions  with  each  other,  as  well  as  to  give  out  information  for  the 
public.  In  this  line  of  thought  he  favored,  also,  the  establishment  of  the 
Concord  School  of  Philosophy,  to  do  a  similar  work  in  the  speculative 
studies ;  and  he  lived  to  see  the  partial  realization  of  what  he  foresaw  in 
this  instance.  In  a  Mathematical  Society  over  which  he  presided  forborne 
years,  each  member  would  bring  something  novel  in  his  own  particular 
branch  of  study;  and  in  the  discussion  which  followed,  it  would  almost 


464  FINKEL'S  SOLUTION   BOOK. 

invariably  appear  that  Professor  Peirce  had,  while  the  paper  was  being^ 
read,  pushed  out  the  author's  methods  to  far  wider  results  than  the 
author  had  dreamed  possible.  The  same  power  of  extending  rapidly  in  his 
own  mind  novel  mathematical  researches  was  exhibited  at  the  sessions  of 
every  scientific  body  at  which  he  chanced  to  be  present.  What  was  quite 
as  admirable  was  the  way  in  which  he  did  it,  giving  the  credit  of  the 
thought  always  to  the  author  of  the  essay  under  discussion.  His  pupils 
thus  frequently  received  credit  for  what  was  in  reality  far  beyond  their 
attainment.  He  robbed  himself  of  fame  in  two  ways:  by  giving  the 
credit  of  his  discoveries  to  those  who  had  merely  suggested  the  line  of 
thought  and  by  neglecting  to  write  out  and  publish  that  which  he  had 
himself  thought  out. 

In  physical  astronomy,  perhaps,  his  greatest  works  were  in  connection 
with  the  planetary  theory,  his  analysis  of  the  Saturnian  system,  his  re- 
searches regarding  the  lunar  theory,  and  the  profound  criticism  of  the  dis- 
covery of  Neptune  following  the  investigations  of  Adams  and  Leverrier. 
At  the  time  of  the  publication  of  his  "System  of  Analytical  Mechanics," 
Professor  Peirce  announced  that  the  volume  would  be  followed  by  three 
others,  entitled  respectively:  "Celestial  Mechanics,"  "Potential  Physics," 
and  "Analytical  Morphology."  These  three  volumes  were  never  pub- 
lished. 

Professor  Peirce,  in  a  paper  read  before  the  American  Association 
for  the  Advancement  of  Science,  in  1849,  showed  in  the  vegetable  world 
the  demonstrable  presence  of  an  intellectual  plan — showed  that  phyllotaxis 
(the  science  of  the  relative  position  of  leaves)  involved  an  algebraic  idea; 
and  this  algebraic  idea  was  subsequently  shown  to  be  the  solution  of  a 
physical  problem. 

The  higher  mathematical  labors  of  so  eminent  a  geometer  must  lie  be- 
yond the  course  of  general  recognition.  Among  the  things  which  give 
him  a  just  claim  to  this  title,  may  be  mentioned:  his  discussion  of  the 
motions  of  two  pendulums  attached  to  a  horizontal  cord;  of  the  motions 
of  a  top;  of  the  fluidity  and  tides  of  Saturn's  rings;  of  the  forms  of 
fluids  enclosed  in  extensible  sacs ;  of  the  motions  of  a  sling ;  of  the  orbits 
of  Uranus,  Neptune,  and  the  comet  of  1843;  of  the  criteria  for  rejecting 
doubtful  observations;  of  a  new  form  of  binary  arithmetic,  of  systems 
of  linear  and  associative  algebra ;  of  various  mechanical  games,  puzzles, 
etc.;  of  various  problems  in  geodesy;  of  the  lunar  tables;  of  the  occul- 
tations  of  the  Pleiades;  etc.  He  adapted  the  epicycles  of  Hipparchus 
to  the  analytical_  forms  of  modern  science;  and  he,  also,  solved  by  a 
system  of  co-ordinates  of  his  own  devising,  several  problems  concerning 
the  involutes  and  evolutes  of  curves,  which  would  probably  have  proved 
impregnable  by  any  other  method  of  mathematical  approach. 

None  of  Professor  Peirce's  labors  lie  farther  above  the  ordinary  reach 
of  thought  than  his  little  lithographed  volume  on  Linear  and  Associative 
Algebra.  In  this  he  discusses  the  nature  of  mathematical  methods;  and 
the  characteristics  which  are  necessary  to  give  novelty  and  unity  to  a  cal- 
culus. Then  he  passes  to  a  description  of  seventy  or  eighty  different 
kinds  of  simple  calculus.  Almost  no  comment  is  given;  but  the  mathe- 
matical reader  discovers,  as  he  proceeds,  that  only  three  species  of  calcu- 
lus, having  each  a  unity  in  itself,  have  been  hitherto  used  to  any  great 
extent,  namely,  —  ordinary  algebra,  differentials,  and  quaternions.  Think 
of  it;  what  a  wonderful  volume  of  prophecy  that  is  which  describes  sev- 
enty or  eighty  species  of  algebra,  any  one  of  which  would  require  genera- 
tion after  generation  of  ordinary  mathematicians  to  develop! 

On  both  sides  of  the  Atlantic,  Professor  Peirce  as  an  author,  was 
highly  esteemed.  His  work  on  analytical  mechanics  was,  at  the  time  of  its 
publication,  regarded  even  in  Germany,  as  the  best  of  its  kind.  As  a 
lecturer,  Professor  Peirce  was  highly  esteemed  in  both  scientific  and  popu- 
lar circles.  It  is  related  that  in  1843,  by  a  series  of  popular  lectures  on 
astronomy,  he  so  excited  the  public  interest  that  the  necessary  funds 


BIOGRAPHY.  465 

•were  immediately  supplied,  for  erecting  an  astronomical  observatory  at 
Harvard  College.  A  remarkable  series  of  lectures  on  "Ideality  in  Science/' 
delivered  by  him  in  1879  before  the  Lowell  Institute  in  Boston,  attracted 
the  general  attention  of  American  thinkers,  on  account  of  the  thoughtful 
consideration  of  the  vexed  question  of  science  and  religion. 

Professor  Peirce  was  a  transcendentalist  in  mathematics,  as  Agassiz 
was  in  zoology ;  and  a  certain  subtile  tie  of  affinity  connected  these  two 
great  men,  however  unlike  they  were  in  their  special  genius.  Alike,  also, 
they  were  in  their  enthusiasm  which  neither  the  piercing  scepticism  of 
Cambridge  could  wither,  nor  declining  years  chill  with  the  frost  of  age. 
The  thing  he  distrusted  was  routine  and  fanatical  method,  whether  new 
or  old ;  for  thought,  salient,  vital,  co-operative  thought,  in  novel  or  in 
ancient  aspects,  he  had  nothing  but  respect  and  furtherance.  Few  men 
could  suggest  more  while  saying  so  little,  or  stimulate  so  much  while 
communicating  next  to  nothing  that  was  tangible  and  comprehensible. 
The  young  man  who  would  learn  the  true  meaning  of  apprehension  as 
distinct  from  comprehension,  should  have  heard  the  professor  lecture,  after 
reciting  to  him.  He  was  always  willing  to  be  esteemed  for  less  than  he 
had  really  accomplished;  and  he  could  join  most  heartily  in  the  praise 
of  others  who  even  owed  their  impulse  to  him.  Modest  and  magnanimous, 
but  not  unobservant,  his  ambition  for  personal  distinction  was  early  and 
easily  satisfied ;  and  he  thus  rid  himself  of  what  is  to  most  men  a  per- 
turbing, and  too  often  an  ignoble,  element  of  discomfort. 

Professor  Peirce  habitually  ascribed  to  his  listener  a  power  of  assimi- 
lation which  the  listener  rarely  possessed.  He  assumed  his  readers  could 
follow  wherever  he  led;  and  this  made  his  lectures  hard  to  follow,  his 
books  brief,  difficult,  and  comprehensive.  When,  however,  his  listeners 
were  students  who  had  previously  attained  some  skill  as  mathematicians 
and  who  had  been  trained  in  his  own  methods,  the  resulting  work  would 
be  of  the  highest  order  of  excellence.  He  was  personally  magnetic  in  his 
presence.  His  pupils  loved  and  revered  him;  and  to  the  young  man,  he 
always  lent  a  helping  hand  in  science.  He  inspired  in  them  a  love  of 
truth  for  its  own  sake. 

His  own  faith  in  Christianity  had  the  simplicity  of  a  child's ;  and 
whatever  radiance  could  emanate  from  a  character  which  combined  the 
greatest  intellectual  attainment  with  the  highest  moral  worth,  that  radi- 
ance cast  its  light  upon  those  who  were  in  his  presence.  "Every  portion 
of  the  material  universe,"  writes  Professor  Peirce,  "is  pervaded  by  the 
same  laws  of  mechanical  action  which  are  incorporated  into  the  very  con- 
stitution of  the  human  mind."  To  him,  then,  the  universe  was  made 
for  the  instruction  of  man.  With  this  belief  he  approached  the  study  of 
natural  phenomena  not  in  the  spirit  of  a  critic,  but  reverently  in  the 
mood  of  a  sympathizing  reader  •  and  the  lesson  he  reads  is :  "There  is 
but  one  God,  and  science  is  the  knowledge  of  Him."  In  his  lectures  and 
teaching  he  showed,  as  he  always  felt  with  adoring  awe,  that  the  mathe- 
matician enters  (as  none  else  can)  into  the  intimate  thought  of  God,  sees 
things  precisely  as  they  are  seen  by  the  Infinite  Mind,  holds  the  scales 
and  compasses  with  which  the  Eternal  Wisdom  built  the  earth  and  meted 
out  the  heavens.  This  consciousness  had  pervaded  his  whole  scientific 
life.  It  was  active  in  his  early  youth,  as  his  coevals  well  remember;  it 
gathered  strength  with  his  years;  and  it  struck  the  ever  recurring  key- 
note in  his  latest  public  utterances. 

Benjamin  Peirce  was  a  devout,  God-fearing  man;  he  was  a  Christian, 
in  the  whole  aim,  tenor,  and  habit  of  his  life.  To  know  Professor  Peirce 
was  simply  to  love  him,  to  admire  him,  and  to  revere  him.  Since  he  was 
conversant  with  the  phases  of  scientific  infidelity,  and  by  no  means  un- 
familiar with  the  historic  grounds  of  scepticism,  it  can  not  be  regarded 
otherwise  than  with  the  profoundest  significance,  that  a  mind  second  to 
none  in  keen  intuition,  in  aesthetic  sensibility,  in  imaginative  fervor,  and 
in  the  capacity  of  close  and  cogent  reasoning,  maintained  through  life  an 


466  FINKEIv'S   SOLUTION   BOOK. 

unshaken  belief  and  trust  in  the  power,  providence,  and  love  of  God,  as> 
beheld  in  his  works,  and  as  incarnate  in  our  Lord  and  Savior.  In  one 
of  his  lectures  on  Ideality  in  Science,  he  said :  "  'Judge  the  tree  by  its 
fruit.'  Is  this  magnificent  display  of  ideality  a  human  delusion?  Or  is 
it  a  divine  record?  The  heavens  and  the  earth  have  spoken  to  declare 
the  glory  of  God.  It  is  not  a  tale  told  by  an  idiot,  signifying  nothing.  It 
is  the  poem  of  an  infinite  imagination,  signifying  immortality." 

In  May,  1880,  Professor  Peirce  began  to  pass  under  the  shadow  of 
the  cloud  of  his  last  illness.  For  some  weeks  there  was  little  serious  fear 
that  it  was  a  shadow  not  destined  to  lift.  He  was  first  confined  to  his 
chamber,  on  the  25th  of  June,  3880;  and  from  that  time,  his  slowly  failing 
condition  was  hardly  relieved  even  by  any  deceptive  appearances  of  im- 
provement. He  died  on  the  morning  of  Wednesday,  October  6,  1880. 
Distinguished  throughout  his  life  by  his  freedom  from  the  usual  abhor- 
rence of  death,  which  he  never  permitted  himself  either  to  mourn  when, 
it  came  to  others,  or  to  dread  for  himself,  he  kept  this  characteristic  temper 
to  the  end,  through  all  the  sad  changes  of  his  trying  illness ;  and,  two 
days  before  he  ceased  to  breathe,  it  struggled  into  utterance  in  a  few 
faintly-whispered  words,  which  expressed  and  earnestly  inculcated  a  cheer- 
ful and  complete  acceptance  of  the  will  of  God  with  regard  to  him. 

The  funeral  took  place  on  Saturday,  October  9,  1880,  at  Appleton 
Chapel,  and  was  the  occasion  of  an  impressive  gathering  of  people  of  great 
and  various  mark.  The  attendance  included  a  very  full  representation  of 
the  various  faculties  and  governing  boards  of  the  University;  a  large 
deputation  of  officers  of  the  United  States  Coast  and  Geodetic  Survey, 
headed  by  the  superintendent  and  the  chief  assistant;  delegations  of  emi- 
nent professors  from  Yale  College  and  the  Johns  Hopkins  University; 
many  members  of  the  class  of  1829;  and  a  great  number  of  other  friends 
of  the  deceased. 

The  pall-bearers  were:  President  Charles  W.  Eliot;  Ex-President 
Thomas  Hill,  Pastor  of  the  First  Parish  Church,  Portland,  Maine;  Capt. 
C.  P.  Patterson,  Superintendent  of  the  United  States  Coast  Survey;  Pro- 
fessor J.  J.  Sylvester,  of  the  Johns  Hopkins  University ;  Hon.  J.  Ingersoll 
Bowditch ;  Professor  Simon  Newcomb,  Superintendent  of  the  American 
Ephemeris  and  Nautical  Almanac;  Dr.  Oliver  Wendell  Holmes;  Pro- 
fessor Joseph  Levering ;  and  Dr.  Merrill  Wyman.  A  beautiful  and  simple 
service  was  conducted  by  the  Rev.  A.  P.  Peabody  and  the  Rev.  James 
Freeman  Clarke. 

In  the  career  of  Professor  Benjamin  Peirce,  America  has  nothing  to 
regret,  but  that  it  is  now  closed;  while  the  American  people  have  much 
to  learn  from  his  long,  useful,  and  honorable  life.  —  By.  F.  P.  Matz. 
From  the  American  Mathematical  Monthly. 

BENJAMIN  PEIRCE. 

For  him  the  Architect  of  all 
Unroofed  our  planet's  starlit  hall ; 
Through  voids  unknown  to  worlds  unseen 
His  clearer  vision  rose  serene. 

With  us  on  earth  he  walked  by  day, 
His  midnight  path  how  far  away ! 
We  knew  him  not  so  well  who  knew 
The  patient  eyes  his  soul  looked  through  ; 

For  who  his  untrod  realm  could  share 
Of  us  that  breathe  this  mortal  air, 
Or  camp  in  that  celestial  tent 
Whose  fringes  gild  our  firmament? 

How  vast  the  workroom  where  he  brought 
The  viewless  implements  of  thought! 
The  wit  how  subtle,  how  profound, 
That  Nature's  tangled  webs  unwound ; 


BIOGRAPHY.  467 


That  through  the  clouded  matrix  saw 

The  chrystal  planes  of  shaping  law, 

Through  these  the  sovereign  skill  that  planned,' 

The  Father's  care,  the  Master's  hand  ! 

To  him  the  wandering  stars  revealed 
The  secrets  in  their  cradle  sealed ; 
The  far-off,  frozen  sphere  that  swings 
Through  ether,  zoned  with  lucid  rings; 

The  orb  that  rolls  in  dim  eclipse 
Wide  wheeling  round  its  long  ellipse,— 
His  name  Urania  writes  with  these 
And  stamps  it  on  her  Pleiades. 

We  knew  him  not?    Ah,  well  we  knew 
The  manly  soul,  so  brave,  so  true, 
The  cheerful  heart  that  conquered  age, 
The  childlike  silver-bearded  sage. 

No  more  his  tireless  thought  explores 
The  azure  sea  with  golden  shores ; 
Rest,  wearied  frame !  the  stars  shall  keep 
A  loving  watch  where  thou  shalt  sleep. 

Farewell !  the  spirit  needs  must  rise, 
So  long  a  tenant  of  the  skies, — 
Rise  to  that  home  all  worlds  above 
Whose  sun  is  God,  whose  light  is  love. 

—Oliver  Wendell  Holmes. 


463  FINKEL'S  SOLUTION  BOOK. 

BIOGRAPHY. 


JAMES  JOSEPH  SYLVESTER,  LL.  D.,  F.  R.  S. 

On  Monday,  March  15,  1897,  in  London,  where,  September  3,  1814,  he 
was  born,  died  the  most  extraordinary  personage  for  half  a  century  in  the 
mathematical  world. 

James  Joseph  Sylvester  was  second  wrangler  at  Cambridge  in  1837. 
When  we  recall  that  Sylvester,  Wm.  Thomson,  Maxwell,  Clifford,  J.  J. 
Thomson  were  all  second  wranglers,  we  involuntarily  wonder  if  any  senior 
wrangler  except  Cayley  can  be  ranked  .with  them. 

Yet  it  was  characteristic  of  Sylvester  that  not  to  have  been  first  was 
always  bitter  to  him. 

The  man  who  beat  him,  Wm.  N.  Griffin,  also  a  Johnion,  afterwards 
a  modest  clergyman,  was  tremendously  impressed  by  Sylvester,  and  hon- 
ored him  in  a  treatise  on  optics  where  he  used  Sylvester's  first  published 
paper,  "Analytical  development  of  Fresnel's  optical  theory  of  crystals," 
Philosophical  Magazine,  1837. 

Sylvester  could  not  be  equally  generous,  and  explicitly  rated  above 
Griffin  the  fourth  wrangler  George  Green,  justly  celebrated,  who  died 
in  1841. 

Sylvester's  second  paper,  "On  the  motion  and  rest  of  fluids/'P/w/o- 
sophical  Magazine,  1838  and  1839,  also  seemed  to  point  to  physics. 

In  1838  he  succeeded  the  Rev.  Wm.  Ritchie  as  professor  of  natural 
philosophy  in  University  College,  London. 

His  unwillingness  to  submit  to  the  religious  tests  then  enforced  at 
Cambridge  and  to  sign  the  39  articles  not  only  debarred  him  from  his 
degree  and  from  competing  for  the  Smith's  prizes,  but,  what  was  far 
worse,  deprived  him  of  the  Fellowship  morally  his  due.  He  keenly  felt  the 
injustice. 

In  his  celebrated  address  at  the  Johns  Hopkins  University  his  denun- 
ciation of  the  narrowness,  bigotry  and  intense  selfishness  exhibited  in  these 
compulsory  creed  tests,  made  a  wonderful  burst  of  oratory.  These  opin- 
ions were  fully  shared  by  De  Morgan,  his  colleague  at  University  Col- 
lege. Copies  I  possess  of  the  five  examination  papers  set  by  Sylvester  at 
the  June  examination,  session  of  1839-40,  show  him  striving  as  a  physicist, 
but  it  was  all  a  false  start.  Even  his  first  paper  shows  he  was  always 
the  Sylvester  we  knew.  To  the  "Index  of  Contents"  he  appends  the  char- 
acteristic note :  "Since  writing  this  index  I  have  made  many  additions 
more  interesting  than  any  of  the  propositions  here  cited,  which  will 
appear  toward  the  conclusion."  Ever  he  is  borne  along  helpless  but 
ecstatic  in  the  ungovernable  flood  of  his  thought. 

A  physical  experiment  never  suggests  itself  to  the  great  mental  experi- 
menter. Cayley  once  asked  for  his  box  of  drawing  instruments.  Sylves- 
ter answered,  "I  never  had  one."  Something  of  this  irksomeness  of  the 
outside  world,  the  world  of  matter,  may  nave  made  him  accept,  in  1841, 
the  professorship  offered  him  in  the  University  of  _  Virginia. 

On  his  way  to  America  he  visited  Rowan  Hamilton  at  Dublin  in  that 
observatory  where  the  maker  of  quaternions  was  as  out  of  place  as  Syl- 
vester himself  would  have  been.  The  Virginians  so  utterly  failed  to  un- 
derstand Sylvester,  his  character,  his  aspirations,  his  powers,  that  the 
Rev.  Dr.  Dabney,  of  Virginia,  has  seriously  assured  me  that  Sylvester' 
was  actually  deficient  in  intellect,  a  sort  of  semi-idiotic  calculating  boy. 
For  the  sake  of  the  contrast,  and  to  show  the  sort  of  civilization  in  which 
this  genius  had  risked  himself,  two  letters  from  Sylvester's  tutors  at  Cam- 
bridge may  here  be  of  interest. 

The  great  Colenso,  Bishop  of  Natal,  previously  Fellow  and  Tutor  of 
St.  John's  College,  writes :  "Having  been  informed  that  my  friend  and 


BIOGRAPHY.  469 

former  pupil,  Mr.  J.  J.  Sylvester,  is  a  candidate  for  the  office  of  professor 
of  mathematics,  I  beg  to  state  my  high  opinion  of  his  character  both  as  a 
mathematician  and  a  gentleman. 

"On  the  former  point,  indeed,  his  degree  of  Second  Wrangler,  at  the 
University  of  Cambridge  would  be,  in  itself,  a  sufficient  testimonial.  But 
I  beg  to  add  that  his  powers  are  of  a  far  higher  order  than  even  that 
degree  would  certify." 

Philip  Kelland,  himself  a  Senior  Wrangler,  and  then  professor  of 
mathematics  in  the  University  of  Edinburgh,  writes :  "I  have  been  re- 
quested to  express  my  opinion  of  the  qualifications  of  Mr.  J.  J.  Sylvester, 
as  a  mathematician. 

"Mr.  Sylvester  was  one  of  my  private  pupils  in  the  University  of  Cam- 
bridge, where  he  took  the  degree  of  Second  Wrangler.  My  opinion  of 
Mr.  Sylvester  then  was  that  in  originality  of  thought  and  acuteness  of 
perception  he  had  never  been  surpassed,  and  I  predicted  for  him  an  emi- 
nent position  among  the  mathematicians  of  Euiope.  My  anticipations 
have  been  verified.  Mr.  Sylvester's  published  papers  manifest  a  depth  and 
originality  which  entitles  them  to  the  high  position  they  occupy  in  the 
field  of  scientific  discovery.  They  prove  him  to  be  a  man  able  to  grapple 
with  the  most  difficult  mathematical  questions  and  are  satisfactory  evidence 
of  the  extent  of  his  attainments  and  the  vigor  of  his  mental  powers." 

The  five  papers  produced  in  this  year,  1841,  before  Sylvester's  depart- 
ure for  Virginia,  show  that  now  his  keynote  is  really  struck.  They 
adumbrate  some  of  his  greatest  discoveries. 

They  are :  "On  the  relation  of  Sturm's  auxiliary  functions  to  the  roots 
of  an  algebraic  equation,"  British  Assoc.  Rep.  (pt.  2),  1841;  "Examples 
of  the  dialytic  method  of  elimination  as  applied  to  ternary  systems  of 
equations,"  Camb.  M.  Jour.  II.,  1841 ;  "On  the  amount  and  distribution 
of  multiplicity  in  an  algebraic  equation,"  Phil.  Mag.  XyiL,  1841 ;  "On 
a  new  and  more  general  theory  of  multiple  roots,"  Phil.  Mag.  XVIIL, 
1841  ;  "On  a  linear  method  of  eliminating  between  double,  treble  and  other 
systems  of  algebraic  equations,"  Phil.  Mag.  XVIIL,  1841 ;  "On  the  dialy- 
tic method  of  elimination,"  Phil.  Mag.  XXL,  Irish  Acad.  Proc.  II. 

This  was  left  behind  in  Ireland,  on  the  way  to  Virginia.  Then  sud- 
denly occurs  a  complete  stoppage  in  this  wonderful  productivity.  Not  one 
paper,  not  one  word,  is  dated  from  the  University  of  Virginia.  Not  until 
1844  does  the  wounded  bird  begin  again  feebly  to  chirp,  and  indeed  it  is  a 
whole  decade  before  the  song  pours  forth  again  with  mellow  vigor  that 
wins  a  waiting  world. 

Disheartening  was  the  whole  experience ;  but  the  final  cause  of  his 
sudden  abandonment  of  the  University  of  Virginia  I  gave  in  an  address 
entitled,  "Original  Research  and  Creative  Authorship  the  Essence  of  Uni- 
versity Teaching,"  printed  in  Science,  N.  S.,  Vol.  I.,  pp.  203-7,  February 
22,  1895. 

On  the  return  to  England  with  heavy  heart  and  dampened  ardor,  he 
takes  up  for  his  support  the  work  of  an  actuary  and  then  begins  the  study 
of  law.  In  1847  we  find  him  at  26  Lincoln's  Inn  Fields,  "eating  his  terms." 
On  November  22,  1850,  he  is  called  to  the  bar  and  practices  conveyancing. 

But  already  in  his  paper  dated  August  12,  1850,  we  meet  the  significant 
names  Boole,  Cayley,  and  harvest  is  at  hand. 

The  very  words  which  must  now  be  used  to  say  what  had  already  hap- 
pened and  what  was  now  to  happen  were  not  then  in  existence.  They 
were  afterward  made  by  Sylvester  and  constitute  in  themselves  a  tre- 
mendous contribution.  As  he  himself  says:  "Names  are,  of  course,  all 
important  to  the  progress  of  thought,  and  the  invention  of  a  really  good 
name,  of  which  the  want,  not  previously  perceived,  is  recognized^  when 
supplied,  as  having  ought  to  be  felt,  is  entitled  to  rank  on  a  level  in  im- 
portance, with  the  discovery  of  a  new  scientific  theory." 

Elsewhere  he  says  of  himself:  "Perhaps  I  may  without  immodesty 
lay  claim  to  the  appellation  of  the  Mathematical  Adam,  as  I  believe  that 


470  FINKEI/S  SOLUTION  BOOK. 

I  have  given  more  names  (passed  into  general  circulation)  to  the  creat- 
ures of  the  mathematical  reason  than  all  the  other  mathematicians  of  the 
age  combined." 

In  one  year,  1851,  Sylvester  created  a  whole  new  continent,  a  new 
world  in  the  universe  of  mathematics.  Demonstration  of  its  creation  is 
given  by  the  Glossary  of  New  Terms  which  he  gives  in  the  Philosophical. 
Transactions,  Vol.  143,  pp.  543-548. 

Says  Dr.  W.  Franz  Meyer  in  his  exceedingly  valuable  Bericht  iiber 
die  Fortschritte  der  projectiven  Invariantentheorie,  the  best  history  of 
the  subject  (1892)  : 

"Als  ausseres  Zeichen  fur  den  Umfang  der  vorgeschrittenen  Entwicke- 
lung  mag  die  ausgedehnte,  grosstenteils  von  Sylvester  selbst  herriihrende 
Terminologie  dienen,  die  sich  am  Ende  seiner  grossen  Abhandlung  iiber 
Sturm'sche  Functionen  (1853)  zusammengestellt  findet." 

Using  then  this  new  language,  let  us  briefly  say  what  had  happened 
in  the  decade  when  Sylvester's  genius  was  suffering  from  its  Virginia 
wound.  The  birthday  of  the  giant  Theory  of  Invariants  is  April  28,  1841, 
the  date  attached  by  George  Boole  to  a  paper  in  the  Cambridge  Mathe- 
matical Journal  where  he  not  only  proved  the  invariantive  property  of  dis- 
criminants generally,  but  also  gave  a  simple  principle  to  form  simultaneous 
invariants  of  a  system  of  two  functions.  The  paper  appeared  in  Novem- 
ber, 1841,  and  shortly  after,  in  February,  1842,  Boole  showed  that  the 
polars  of  a  form  lead  to  a  broad  class  of  covariants.  Here  he  extended 
the  results  of  the  first  article  to  more  than  two  Forms.  Boole's  papers 
led  Cayley,  nearly  three  years  later  (1845),  to  propose  to  himself  the 
problem  to  determine  a  priori  what  functions  of  the  coefficients  of  an 
equation  possess  this  property  of  invariance,  and  he  discovered  its  pos- 
session by  other  functions  besides  discriminants,  for  example  the  quad- 
rinvariants  of  binary  quantics,  and  in  particular  the  invariant  S  of  a 
quartic. 

Boole  next  discovered  the  other  invariant  T  of  a  quartic  and  the  ex- 
pression of  the  discriminant  in  terms  of  S  and  T.  Cayley  next  (1846) 
published  a  symbolic  method  of  finding  invariants.  Early  in  1851  Boole 
reproduced,  with  additions,  his  paper  on  Linear  Transformations;  then 
at  last  began  Sylvester.  He  always  mourned  what  he  called  "the  years 
he  lost  fighting  the  world" ;  but,  after  all,  it  was  he  who  made  the 
Theory  of  Invariants. 

Says  Meyer :  "sehen  wir  in  dem  Cyklus  Sylvester'scher  Publicationen 
(1851-1854)  bereits  die  Grundzuge  einer  allegemeinen  Theorie  erstehen, 
welche  die  Elemente  von  den  verschiedenartigsten  Zweigen  der  spateren 
Disciplin  umfasst."  "Sylvester  beginnt  damit,  die  Ergebnisse  seiner  Vor- 
ganger  unter  einem  einzigen  Gesichtspunkte  zu  vereinigen." 

With  deepest  foresight  Sylvester  introduced,  together  with  the  original 
variables,  those  dual  to  them,  and  created  the  theory  of  contravariants  and 
intermediate  forms.  He  introduced,  with  many  other  processes  for  pro- 
ducing invariantive  forms,  the  principle  of  mutual  differentiation. 

Hilbert  attributes  the  sudden  growth  of  the  theory  to  these  processes 
for  producing  and  handling  invariantive  creatures.  "Die  Theorie  dieser 
Gebilde  erhob  sich,  von  speciellen  Aufgaben  ausgehend,  rasch  zu  grosser 
Allgemeinheit  —  dank  vor  Allem  dem  Umstande,  dass  es  gelang,  eine 
Reihe  von  besonderen  der  Invariantentheorie  eigenthumlichen  Prozessen 
zu  entdecken,  deren  Anwendung  die  Aufstellung  und  Behandlung  invari- 
anter  Bildungen  betrachtlich  erleichterte." 

"Was  die  Theorie  der  algebraischen  Invarianten  anbetrifft  so  sind  die 
ersten  Begriinder  derselben,  Cayley  und  Sylvester,  zugleich  auch  als  die 
Vertreter  der  naiven  Periode  anzusehen :  an  der  Aufstellung  der  einfach- 
sten  Invariantenbildungen  und  an  den  eleganten  Anwendungen  auf  die 
Auflosung  der  Gleichungen  der  ersten  4  Grade  hatten  sie  die  unmittelbare 
Freude  der  ersten  Entdeckung."  It  was  Sylvester  alone  who  created  the 


BIOGRAPHY.  471 

theory  of  canonic  forms  and  proceeded  to  apply  it  with  astonishing  power. 
What  marvelous  mass  of  brand  new  being  he  now  brought  forth ! 

Moreover  he  trumpeted  abroad  the  eruption.  He  called  for  communi- 
cations to  himself  in  English,  French,  Italian,  Latin  or  German,  so  only 
the  "Latin  character"  were  used. 

From  1851  to  1854  he  produces  forty-six  different  memoirs.  Then 
comes  a  dead  silence  of  a  whole  year,  broken  in  1856  by  a  feeble  chirp 
called  "A  Trifle  on  Projectiles." 

What  has  happened?  Some  more  "fighting  the  world."  Sylvester 
declared  himself  a  candidate  for  the  vacant  professorship  of  geometry  in 
Gresham  College,  delivered  a  probationary  lecture  on  the  4th  of  Decem- 
ber, 1854,  and  was  ignominiously  "turned  down."  Let  us  save  a  couple  of 
sentences  from  this  lecture : 

"He  who  would  know  what  geometry  is  must  venture  boldly  into  its 
depths  and  learn  to  think  and  feel  as  a  geometer.  I  believe  that  it  is 
impossible  to  do  this,  to  study  geometry  as  it  admits  of  being  studied,  and 
I  am  conscious  it  can  be  taught,  without  finding  the  reasoning  invigorated, 
the  invention  quickened,  the  sentiment  of  the  orderly  and  beautiful  awak- 
ened and  enhanced,  and  reverence  for  truth,  the  foundation  of  all  integrity 
of  character,  converted  into  a  fixed  principle  of  the  mental  and  moral 
constitution,  according  to  the  old  and  expressive  adage  'abeunt  studia  in 
mores/  " 

But  this  silent  year  concealed  still  another  stunning  blow  of  precisely 
the  same  sort,  as  bears  witness  the  f611owing  letter  from  Lord  Brougham 
to  The  Lord  Panmure: 

"  BROUGHAM, 

PRIVATE.  28  Aug.  1855. 

MY  DEAR  P. 

My  learned  excellent  friend  and  brother  mathematician  Mr.  Sylvester  is  again  a 
candidate  for  the  professorship  at  Woolwich  on  the  death  of  Mr.  O'Brian  who  carried  it 
against  him  last  year. 

I  entreat  once  more  your  favorable  consideration  of  this  eminent  man  who  has  al- 
ready to  thank  you  for  your  great  kindness. 

Yours  sincerely, 

H.  BROUGHAM. 

On  this  third  trial,  backed  by  such  an  array  of  credentials  as  no  man 
ever  presented  before,  he  barely  scraped  through,  was  appointed  professor 
of  mathematics  at  the  Royal  Military  Academy,  and  served  at  Woolwich 
exactly  14  years,  10  months,  and  15  days. 

A  single  sentence  of  his  will  best  express  his  greatest  achievement 
there  and  his  manner  of  exit  thence : 

"If  Her  most  Gracious  Majesty  should  ever  be  moved  to  recognize 
the  palmary  exploit  of  the  writer  of  this  note  in  the  field  of  English  science 
as  having  been  the  one  successfully  to  resolve  a  question  and  conquer  an 
algebraical  difficulty  which  had  exercised  in  vain  for  two  centuries  past, 
since  the  time  of  Newton,  the  highest  mathematical  intellects  in  Europe 
(Euler,  Lagrange,  Maclaurin,  Waring  among  the  number),  by  conferring 
upon  him  some  honorary  distinction  in  commemoration  of  the  deed,  he 
will  crave  the  privilege  of  being  allowed  to  enter  the  royal  presence,  not 
covered,  like  De  Courcy,  but  barefooted,  with  rope  around  his  waist,  and 
a  goose-quill  behind  his  ear,  in  token  of  repentant  humility,  and  as  an 
emblem  of  convicted  simplicity  in  haying  once  supposed  that  on  such 
kind  of  success  he  could  found  any  additional  title  to  receive  fair  and  just 
consideration  at  the  hands  of  Her  Majesty's  Government  when  quitting 
his  appointment  as  public  professor  at  Woolwich  under  the  coercive  opera- 
tion of  a  non-Parliamentary  retrospective  and  utterly  unprecedented  War 
Office  enactment."  Athenaeum  Club,  January  31,  1871.  Of  course  this 
means  a  row  of  barren  years,  1870,  1871,  1872,  1873. 

The  fortunate  accident  of  a  visit  paid  Sylvester  in  the  autumn  of  1873 
by  Pafnuti  Lvovich  Chebyshev,  of  the  University  of  St.  Petersburg,  re- 
awakened our  genius  to  produce  in  a  single  burst  of  enthusiasm  a  new- 
branch  of  science. 


472  FINKEVS  SOLUTION  BOOK. 

On  Friday  evening,  January  23,  1874,  Sylvester  delivered  at  the  Royal 
Institution  a  lecture  entitled  "On  Recent  Discoveries  in  Mechanical  Con- 
version of  Motion,"  whose  ideas,  carried  on  by  two  of  his  hearers,  H. 
Hart  and  A.  B.  Kempe,  have  made  themselves  a  permanent  place  even 
in  the  elements  of  geometry  and  kinematics.  A  synopsis  of  this  lecture 
was  published,  but  so  curtailed  and  twisted  into  the  third  person  that  the 
life  and  flavor  are  quite  gone  from  it.  I  possess  the  unique  manuscript 
of  this  epoch-making  lecture  as  actually  delivered.  A  few  sentences  will 
show  how  characteristic  and  inimitable  was  the  original  form: 

"The  air  of  Russia  seems  no  less  favorable  to  mathematical  acumen 
than  to  a  genius  for  fable  and  song.  Lobacheffsky,  the  first  to  mitigate 
the  severity  of  the  Euclidean  code  and  to  beat  down  the  bars  of  a  supposed 
adamantine  necessity,  was  born  (a  Russian  of  Russians),  in  the  govern- 
ment of  Nijni  Novgorod;  Tchebicheff  [Chebyshev],  the  prince  and  con- 
queror of  prime  numbers,  able  to  cope  with  their  refractory  character  and 
to  confine  the  stream  of  their  erratic  flow,  their  progression,  within  alge- 
braic limits,  in  the  adjacent  circumscription  of  Moscow;  and  our  own 
Cayley  was  cradled  amidst  the  snows  of  St.  Petersburg."  [Sylvester  him- 
self contracted  Chebyshev's  limits  for  the  distribution  of  primes.]  "I 
think  I  may  fairly  affirm  that  a  simple  direct  solution  of  the  problem  of 
the  duplication  of  the  cube  by  mechanical  means  was  never  accomplished 
down  to  this  day.  .1  will  not  say  but  that,  by  a  merciful  interpretation  of 
his  oracle,  Apollo  may  have  put  up  with  the  solution  which  the  ancient 
geometers  obtained  by  means  of  drawing  two  parabolic  curves ;  but  of 
this  I  feel  assured  that  had  I  been  then  alive,  and  could  have  shown  my 
solution,  which  I  am  about  to  exhibit  to  you,  Apollo  would  have  leaped 
for  joy  and  danced  (like  David  before  the  ark),  with  my  triple  cell  in 
hand,  in  place  of  his  lyre,  before  his  own  duplicated  altar." 

That  in  the  very  next  year  Sylvester  was  taking  a  more  active  part 
than  has  hitherto  been  known  in  the  organization  of  the  incipient  Johns 
Hopkins  University  is  seen  from  the  following  letter  to  him  in  London 
from  the  great  Joseph  Henry : 

SMITHSONIAN  INSTITUTION, 

August  25, 1875. 
MY  DEAR  SIR: 

Your  letter  of  the  13th  inst.  has  just  been  received  and  in  reply  I  have  to  say  that  I 
have  written  to  President  Gilman  of  the  Hopkins  Universit}'  giving  my  views  as  to  what 
it  ought  to  be  and  have  stated  that  if  properly  managed  it  may  do  more  for  the  advance 
of  literature  and  science  in  this  country  than  any  other  institution  ever  established ;  it 
is  entirely  independent  of  public  favor  and  may  lead  instead  of  following  popular 
opinion. 

I  have  advised  that  liberal  salaries  be  paid  to  the  occupants  of  the  principal  chairs 
and  that  to  fill  them  the  best  men  in  the  world  who  can  be  obtained  should  be  secured. 

I  have  mentioned  your  name  prominently  as  one  of  the  very  first  mathematicians 
of  the  day  ;  what  the  result  will  be,  however,  I  can  not  say. 

The  Trustees  are  all  citizens  of  Baltimore  and  among  them  I  have  some  personal 
friends  ;  the  President,  Mr.  Gilman,  and  one  of  them,  came  to  Washington  a  few  weeks 
ago  to  get  from  me  any  suggestions  that  I  might  have  to  offer. 

It  is  to  be  regretted  that  in  this  country  the  Trustees,  who  control  the  management 
of  bequests  of  this  character,  think  it  important  to  produce  a  palpable  manifestation  of 
the  institution  to  be  established  by  spending  a  large  amount  of  the  bequest  in  archi- 
tectural displays.    Against  this  custotn  I  have  protested  and  have  asserted  that  if  the 
E roper  men  and  necessary  implements  of  instruction  are  provided,  the  teaching  may 
e  done  in  log  cabins. 

It  would  give  me  great  pleasure  to  have  you  again  as  my  guest,  and  I  will  do  what  I 
can  to  secure  your  election.  Very  truly  your  friend, 

JOSEPH  HENRY. 
We  know  the  result. 

Sylvester  was  offered  the  place;  demanded  a  higher  salary;  won; 
came. 

I  was  his  first  pupil,  his  first  class,  and  he  always  insisted  that  it  was 
I  who  brought  him  back  to  the  Theory  of  Invariative  Forms.  In  a  letter 
to  me  of  September  24,  1882,  he  writes :  "Nor  can  I  ever  be  oblivious  of 
the  advantage  which  I  derived  from  your  well-grounded  persistence  in 
inducing  me  to  lecture  on  the  Modern  Algebra,  which  had  the  effect  of 
bringing  my  mind  back  to  this  subject,  from  which  it  had  for  some  time 


BIOGRAPHY.  47a 

previously  been  withdrawn,  and  in  which  I  have  been  laboring,  with  a 
success  which  has  considerably  exceeded  my  anticipations,  ever  since." 

He  made  this  same  statement  at  greater  length  in  his  celebrated  ad- 
dress at  the  Johns  Hopkins  on  February  22,  1877:  "At  this  moment  I 
happen  to  be  engaged  in  a  research  of  fascinating  interest  to  myself,  and 
which,  if  the  day  only  responds  to  the  promise  of  its  dawn,  will  meet,  I 
believe,  a  sympathetic  response  from  the  professors  of  our  divine  alge- 
braical art  wherever  scattered  through  the  world. 

"There  are  things  called  Algebraical  Forms ;  Professor  Cayley  calls 
them  Quantics.  These  are  not,  properly  speaking,  Geometrical  Forms,  al- 
though capable,  to  some  extent,  of  being  embodied  in  them,  but  rather 
schemes  of  processes,  or  of  operations  for  forming,  for  calling  into  exist- 
ence, as  it  were,  algebraic  quantities. 

"To  every  such  Quantic  is  associated  an  infinite  variety  oi  other  forms 
that  may  be  regarded  as  engendered  from  and  floating,  like  an  atmosphere, 
around  it ;  but  infinite  in  number  as  are  these  derived  existences,  these 
emanations  from  the  parent  form,  it  is  found  that  they  admit  of  being 
obtained  by  composition,  by  mixture,  so  to  say,  of  a  certain  limited  num- 
ber of  fundamental  forms,  standard  rays,  as  they  might  be  termed,  in  the 
Algebraic  Spectrum  of  the  Quantic  to  which  they  belong;  and,  as  it  is  a 
leading  pursuit  of  the  physicists  of  the  present  day  to  ascertain  the  fixed 
lines  in  the  spectrum  of  every  chemical  substance,  so  it  is  the  aim  and 
object  qf  a  great  school  of  mathematicians  to  make  out  the  fundamental 
derived  forms,  the  Covariants  and  Invariants,  as  they  are  called,  of  these 
Quantics. 

"This  is  the  kind  of  investigation  in  which  I  have,  for  the  last  month 
or  two,  been  immersed,  and  which  I  entertain  great  hopes  of  bringing 
to  a  successful  issue. 

"Why  do  I  mention  it  here?  It  is  to  illustrate  my  opinion  as  to  the 
invaluable  aid  of  teaching  to  the  teacher,  in  throwing  him  back  upon  his 
own  thoughts  and  leading  him  to  evolve  new  results  from  ideas  that 
would  have  otherwise  remained  passive  or  dormant  in  his  mind. 

"But  for  the  persistence  of  a  student  of  this  university  in  urging  upon 
me  his  desire  to  study  with  me  the  modern  algebra  I  should  nevker  have 
been  led  into  this  investigation ;  and  the  new  facts  and  principles  which 
I  have  discovered  in  regard  to  it  (important  facts,  I  believe)  would,  so 
far  as  I  am  concerned,  have  remained  still  hidden  in  the  womb  of  time. 
In  vain  I  represented  to  this  inquisitive  student  that  he  would  do  better 
to  take  up  some  other  subject  lying  less  off  the  beaten  track  of  study, 
such  as  the  higher  parts  of  the  Calculus  or  Elliptic  Functions,  or  the  theory 
of  Substitutions,  or  I  wot  not  what  besides.  He  stuck  with  perfect  re- 
spectfulness, but  with  invincible  pertinacity,  to  his  point.  He  would  have 
the  New  Algebra  (Heaven  knows  where  he  had  heard  about  it,  for  it  is 
almost  unknown  on  this  continent),  that  or  nothing.  I  was  obliged  to 
yield,  and  what  was  the  consequence?  In  trying  to  throw  light  upon  an 
obscure  explanation  in  our  text-book  my  brain  took  fire ;  I  plunged  with 
requickened  zeal  into  a  subject  which  I  had  for  years^  abandoned,  and 
found  food  for  thoughts  which  have  engaged  my  attention  for  a  consider- 
able time  past,  and  will  probably  occupy  all  my  powers  of  contemplation 
advantageously  for  several  months  to  come." 

Another  specific  instance  of  the  same  thing  he  mentions  in  his  paper, 
"Proof  of  the  Hitherto  Undemonstrated  Fundamental  Theorem  of  Inyari- 
ants,"  dated  November  13,  1877: 

"I  am  about  to  demonstrate  a  theorem  which  has  been  waiting  proof 
for  the  last  quarter  of  a  century  and  upwards.  It  is  the  more  necessary 
that  this  should  be  done,  because  the  theorem  has  been  supposed  to  lead 
to  false  conclusions,  and  its  correctness  has  consequently  been  impugned. 
Thus  in  Professor  Faa  de  Bruno's  valuable  Theorie  des  formes  binaires, 
Turin,  1876,  at  the  foot  of  page  150  occurs  the  following  passage:  ''Cela 
suppose  essentiellement  que  les  equations  de  condition  soient  toutes  inde'- 


474  FlNKEIv'S  SOLUTION  BOOK. 

pendantes  entr'elles,  ce  qui  n'est  pas  toujours  le  cas,  ainsi  qu'il  resulte 
des  recherches  du  Professor  Gordan  sur  les  nombres  des  covariants  des 
formes  quintique  et  sextique." 

The  reader  is  cautioned  against  supposing  that  the  consequence  alleged 
above  does  result  from  Gordan's  researches,  which  are  indubitably  correct. 
This  supposed  consequence  must  have  arisen  from  a  misapprehension,  on 
the  part  of  M.  de  Bruno,  of  the  nature  of  Professor  Cayley's  rectification 
of  the  error  of  reasoning  contained  in  his  second  memoir  on  Quantics, 
which  had  led  to  results  discordant  with  Gordan's.  Thus  error  breeds 
error,  unless  and  until  the  pernicious  brood  is  stamped  out  for  good  and 
all  under  the  iron  heel  of  rigid  demonstration.  In  the  early  part  of  this 
year  Mr.  Halsted,  a  fellow  of  Johns  Hopkins  University,  called  my  atten- 
tion to  this  passage  in  M.  de  Bruno's  book;  and  all  I  could  say  in  reply 
was  that  'the  extrinsic  evidence  in  support  of  the  independence  of  the  equa- 
tions which  had  been  impugned  rendered  it  in  my  mind  as  certain  as  any 
fact  in  nature  could  be,  but  that  to  reduce  it  to  an  exact  demonstration 
transcended,  I  thought,  the  powers  of  the  human  understanding.'  " 

In  1883  Sylvester  was  made  Savilian  professor  of  geometry  at  Oxford, 
the  first  Cambridge  man  so  honored  since  the  appointment  of  W.allis  in 
1649. 

To  greet  the  new  environment,  he  created  a  new  subject  for  his  re- 
searches —  Reciprocants,  which  has  inspired,  among  others,  J.  Hammond, 
of  Oxford;  McMahon,  of  Woolwich;  A.  R.  Forsyth,  of  Cambridge; 
Leudesdorf,  Elliott  and  Halphen. 

Sylvester  never  solved  exercise  problems  such  as  are  proposed  in  the 
Educational  Times,  though  he  made  them  all  his  life  long  down  to  his 
latest  years.  For  example,  unsolved  problems  by  him  will  be  found  even 
in  Vol.  LXIL  and  Vol.  LXIII.  of  the  Educational  Times  reprints  (1395). 
If  at  the  time  of  meeting  his  own  problem  he  met  also  a  neat  solution 
he  would  communicate  them  together,  but  he  never  solved  any.  In  the 
meagre  notices  that  have  been  given  of  Sylvester  the  strangest  errors 
abound.  Thus  C.  S.  Pierce,  in  the  Post,  March  16th,  speaks  of  his 
accepting,  "with  much  diffidence,"  a  word  whose  meaning  he  never  knew; 
and  gives  1862  as  the  date  of  his  retirement  from  Woolwich,  which  is 
eight  years  wrong,  as  this  forced  retirement  was  July  31,  1870,  after  his 
55th  birthday.  Cajori,  in  his  inadequate  account  (History  of  Mathematics, 

&326),  puts  the  studying  of  law  before  the  professorship  at  University 
^  allege  and  the  professorship  at  the  University  of  Virginia,  both  of  which 
it  followed.  Effect  must  follow  cause.  And  strange,  that  of  the  few 
things  he  ascribes  to  Sylvester,  he  should  have  hit  upon  something  not 
his,  "the  discovery  of  the  partial  differential  equations  satisfied  by  the 
invariants  and  covariants  of  binary  quantics."  But  Sylvester  has  ex- 
plicitly said  in  Section  VI.  of  his  "Calculus  of  Forms" :  "I  alluded  to 
the  partial  differential  equations  by  which  every  invariant  may  be  de- 
fined. M.  Aronhold,  as  I  collect  from  private  information,  was  the  first 
to  think  of  the  application  of  this  method  to  the  subject;  but  it  was  Mr. 
Cayley  who  communicated  to  me  the  equations  which  define  the  invariants 
of  functions  of  two  variables." 

Surely  he  needs  nothing  but  his  very  own,  this  marvellous  man  who 
gave  so  lavishly  to  every  one  devoted  to  mathematics,  or,  indeed,  to  the 
highest  advance  of  human  thought  in  any  form.  —  By  George  Bruce  Hal- 
sted. From  the  American  Mathematical  Monthly. 


BIOGRAPHY.  475 


BIOGRAPHY. 


ARTHUR  CAYLEY. 

Arthur  Cayley  was  born  at  Richmond  in  Surrey,  England,  August 
the  16th,  1821.  His  father,  Henry  Cayley,  was  descended  from  the  Cay- 
leys  of  Brompton,  in  Yorkshire,  but  was  at  the  time  a  merchant  of  St. 
Petersburg  where  he  had  married  a  Russian  lady.  In  1829  his  parents 
took  up  their  permanent  residence  at  Blackheath  in  England ;  and  Arthur 
was  there  educated  at  a  private  school  for  four  years.  At  the  age  of  14 
he  was  sent  to  King's  College  School,  London;  and  the  master  of  that 
school  having  observed  the  promise  of  a  mathematical  genius  advised 
the  father  to  educate  his  son  not  for  his  own  business,  but  to  enter  the 
University  of  Cambridge. 

In  1838  Arthur  Cayley  entered  Trinity  College,  Cambridge,  at  the 
rather  early  age  of  17.  Throughout  his  undergraduate  course  he  was 
first  at  his  college  examinations  by  an  enormous  interval,  and  he  finished 
his  undergraduate  career  in  1842  by  carrying  off  the  two  highest  honors, 
namely,  the  first  place,  or  Senior  Wrangler,  in  the  Mathematical  Tripos, 
and  the  first  prize  in  the  competition  for  the  Smith  Prizes.  Immediately 
elected  a  Fellow  of  his  College,  he  continued  to  reside  at  Cambridge  for 
several  years,  during  which  time  he  lectured  on  mathematics,  and  also 
contributed  papers  to  the  Cambridge  Mathematical  Journal.  His  first 
contribution  to  that  Journal  was  made,  when  he  was  an  undergraduate, 
in  1841. 

At  that  time  it  was  necessary  for  a  Fellow  to  take  Holy  Orders,  or 
else  resign  the  fellowship  at  the  end  of  seven  years.  Mr.  Cayley  chose  the 
latter  alternative,  and  became  by  profession  a  conveyance  in  Lincoln's  Inn, 
London.  He  followed  that  profession  for  14  years  with  conspicuous  ability 
and  success,  and  at  the  same  time  made  many  of  his  most  important  con- 
tributions to  mathematical  science. 

About  1861  the  Lucasian  professorship  of  mathematics  at  Cambridge 
—  the  chair  made  illustrious  by  Sir  Isaac  Newton  —  fell  vacant;  it  was 
filled  by  G.  G.  Stokes,  already  eminent  for  his  work  in  mathematical 
physics,  and  Senior  Wrangler  the  year  before  Cayley.  However,  it  was 
felt  desirable  to  secure  Cayley  also,  and  for  this  purpose  the  -Sadlerian 
professorship  of  mathematics  was  created,  which  resulted  in  Cayley  marry- 
ing and  settling  down  at  Cambridge,  in  1863. 

The  duties  of  the  Sadlerian  professor  were  defined  as  follows :  "to 
explain  and  teach  the  principles  of  pure  mathematics,  and  to  apply  himself 
to  the  advancement  of  the  science."  In  carrying  out  the  former  part  of 
the  duties  Professor  Cayley  did  not  give  the  same  course  of  lectures  year 
after  year,  but  each  year  took  for  his  subject  that  of  the  memoir  on 
which  he  was  engaged.  As  a  consequence  his  students  were  few,  for 
advanced  work  of  that  kind  did  not  pay  in  the  great  mathematical  exam- 
ination. How  well  he  carried  out  the  second  part  of  the  duties  may  be 
inferred  from  the  fact  that  the  Royal  Society  Catalogue  of  Scientific 
Papers  enumerates  430  memoirs  contributed  by  him  between  the  years 
1863  and  1883,  making  a  total  up  to  the  latter  date  of  724.  As  he  con- 
tinued active  to  the  last,  it  is  probable  that  the  grand  total  of  his  papers 
does  not  fall  short  of  1000.  Some  of  his  most  celebrated  contributions 
are:  Chapters  in  the  Analytical  Geometry  of  (n)  Dimensions,  On  the 
theory  of  Determinants,  On  the  theory  of  linear  transformations,  Ten 
Memoirs  on  Quantics,  Memoir  on  the  theory  of  Matrices,  Memoirs  on 
Skew  Surfaces,  otherwise  Scrolls,  On  the  Motion  of  Rotation  of  a  Solid 
Body,  On  the  triple  tangent  planes  of  surfaces  of  the  third  order.  Several 
of  his  achievements  are  elegantly  referred  to  in  a  poem  written  by  his 


476  FINKEI/S  SOLUTION  BOOK. 

colleague  Clerk  Maxwell   in   1874,   and  addressed  to  the   Committee  of 
subscribers  who  had  charge  of  the  Cayley  Portrait  Fund : 

O  wretched  race  of  men,  to  space  confined  ! 

What  honor  can  ye  pay  to  him  whose  mind 

To  that  which  lies  beyond  hath  penetrated? 

The  symbols  he  hath  formed  shall  sound  his  praise, 

And  lead  him  on  through  unimagiued  ways 

To  conquests  new,  in  worlds  not  yet  created. 

First,  ye  Determinants,  in  order  row 

And  massive  column  ranged,  before  him  go, 

To  form  a  phalanx  for  his  safe  protection, 

Ye  powers  of  the  nth  root  of — 1 ! 

Around  his  head  in  endless  cycles  run» 

As  unembodied  spirits  of  direction. 

And  you,  ye  undevelopable  scrolls ! 

Above  the  host  wave  your  emblazoned  rolls, 

Ruled  for  the  record  of  his  bright  inventions. 

Ye  cubic  surfaces !  by  threes  and  nines 

Draw  round  his  camp  your  seven  and  twenty  lines 

The  seal  of  Solomon  in  three  dimensions. 

March  on,  symbolic  host !  with  step  sublime, 
Up  to  the  naming  bounds  of  Space  and  Time ! 
There  pause,  until  by  Dickenson  depicted, 
In  two  dimensions,  we  the  form  may  trace 
Of  him  whose  soul,  top  large  for  vulgar  space, 
In  n  dimensions  flourished  unrestricted. 

The  portrait  was  presented  to  Trinity  College,  and  now  adorns  their 
Hall.  He  is  represented  as  seated  at  a  desk,  with  quill  in  hand,  and  think- 
ing out  intently  some  mathematical  idea. 

But  mathematical  science  was  advanced  by  Professor  Cayley  in  yet 
another  way.  By  his  immense  learning,  his  impartial  judgment,  and  his 
friendly  sympathy  with  other  workers,  he  was  eminently  qualified  to  act 
as  a  referee  on  mathematical  papers  contributed  to  the  various  societies. 
Of  this  kind  of  work  he  did  a  large  amount,  and  of  his  kindliness  to  young 
investigators  I  can  speak  from  personal  experience.  Several  papers  which 
I  read  before  the  Royal  Society  of  Edinburgh  were  referred  to  him,  and 
he  recommended  their  publication.  Some  time  after  I  attended  a  meeting 
of  the  Mathematical  Society  of  London,  but  the  friend  who  would  have 
introduced  me  could  not  be  present.  Professor  Cayley  was  present,  and 
on  finding  out  who  I  was,  gave  me  a  cordial  handshake,  and  referred  in 
the  kindest  terms  to  the  papers  he  had  read.  He  was  a  cosmopolitan 
spirit,  delighting  only  in  the  truth,  and  friendly  to  all  seekers  after  the 
truth. 

Among  Cayley's  papers  there  are  several  on  a  "Question  in  the 
Theory  of  Probabilities."  The  question  was  propounded  by  Boole,  and  he 
applied  to  its  solution  the  general  method  of  "The  Laws  of  Thought."  It 
was  afterwards  discussed  by  Wilbraham,  Cayley  and  others  in  the  Philo- 
sophical Magazine.  My  attention  was  drawn  to  the  question  when  writ- 
ing the  Principles  of  the  Algebra  of  Logic,  and  I  ventured  to  contribute 
my  idea  of  the  question  to  the  Educational  Times.  On  mentioning  the 
matter  to  Professor  Kelland,  he  intimated  pretty  plainly  that  the  discus- 
sion had  been  closed  by  Professor  Cayley,  and  that  it  was  temerity  on  my 
part  to  write  anything  on  the  subject.  But  the  great  mathematician  did 
not  think  so;  he  wrote  me  a  letter  discussing  the  question  and  my, par- 
ticular way  of  viewing  it,  as  well  as  the  fundamental  ideas  in  which  I 
differed  from  Boole. 

In  1882  he  received  a  flattering  invitation  from  the  trustees  of  the 
Johns  Hopkins  University  to  deliver  a  course  of  lectures  on  some  subject 
in  advanced  mathematics.  He  chose  as  his  subject  the  Elliptic  and 
Abelian  functions ;  and  the  impression  which  his  presence  created  has 
been  well  described  by  Dr.  Matz  in  his  brief  notice  in  the  January  number 
of  the  MONTHLY. 


BIOGRAPHY.  477 

Next  year  he  was  president  of  the  British  Association  at  the  South- 
port  meeting.  In  his  address  he  spoke  of  the  foundations  of  mathematics, 
reviewed  the  more  important  theories,  traced  the  connection  of  pure  with 
applied  mathematics,  and  gave  an  outline  of  the  vast  extent  of  Modern 
Mathematics. 

He  regarded  the  complex  number  a  -f-  bi  as  the  fundamental  quantity 
of  mathematical  analysis,  and  considered  that  with  such  a  basis,  algebra 
was  a  complete  and  bounded  science,  in  which  no  further  imaginary  sym- 
bols could  spring  up.  It  is  the  more  remarkable  that  he  held  such  a 
view,  when  we  consider  that  early  in  his  career  he  made  a  notable  con- 
tribution to  space  analysis.  Starting  from  Rodrigues'  formulae  for  the 
rotation  of  a  solid  body,  he  arrived  at  the  quaternion  formula,  and  was 
anticipated  by  Hamilton  only  by  a  few  months.  But  Cayley  took  a 
Cartesian  view  of  analysis  to  the  last,  as  is  evident  from  the  chapter 
which  he  contributed  to  Tait's  Treatise  on  Quaternions.  His  aim  there  is 
to  give  an  analytical  theory  of  quaternions.  Hamilton's  aim  on  the  other 
hand  was  to  give  a  quaternionic  theory  of  analysis.  The  difference  is 
brought  out  still  more  strikingly  in  a  paper  printed  in  the  last  number  of 
the  Proceedings  of  the  Royal  Society  of  Edinburgh. 

In  1889  the  Cambridge  University  Press  commenced  the  re-publication 
of  his  mathematical  papers  in  a  collected  form.  It  was  calculated  that 
they  would  occupy  10  quarto  volumes ;  12  volumes  have  already  appeared ; 
and  it  is  believed  that  13  volumes  will  be  required.  No  mathematician  has 
ever  had  his  works  printed  in  a  more  handsome  manner.  In  addition  he 
is  the  author  of  a  separate  work  on  Elliptic  Functions. 

Space  fails  to  enumerate  the  honors  which  he  received  from  Univer- 
sities and  Scientific  Academies  both  of  the  Old  and  of  the  New  World. 
But  we  may  mention  specially,  that  from  the  Royal  Society  he  received 
a  Royal  Medal  and  a  Copley  Medal;  from  the  Mathematical  Society  of 
London  the  first  DeMorgan  Medal;  and  at  the  instance  of  the  President 
and  Members  of  the  French  Academy  he  was  made  an  Officer  of  the  Legion 
of  Honour. 

On  the  26th  of  January  he  died  at  Cambridge.  His  body  was  laid  to 
rest  in  Mill  Road  Cemetery  in  the  presence  of  official  representatives  from 
foreign  countries  and  many  of  the  most  illustrious  philosophers  of  Eng- 
land. His  spirit  still  speaks  to  us  from  his  works,  and  will  continue  to 
speak  to  many  succeeding  generations.  —  By  Dr.  Alexander  Macfarlane. 
From  the  American  Mathematical  Monthly,  Vol.  II.,  No.  4.  In  the  same 
number  is  also*  an  interesting  biography  of  Cayley,  by  Dr.  George  Bruce 
Halsted. 


478 


FINKEL'S    SOLUTION    BOOK. 


TABLE   I. — Functions  of  n  and  e. 


n  =  3-1415926 
7r2=  9-8696044 
7T3=31-0062761 
*/*==  1-7724539 
log107T=  1-4971499 
loge  ?r=  0-6679358 

jr-i=       -3183099 
7T-2=      -1013212 
7T-8=      -0322515 
200s7  -7-^=63*  -6619772 
1800-7-7r=57°-2957795 
=206264/x-8 

e    =2-71828183 
*2  =7-38905611 
^-1=0-3678794 
^-2=0-1353353 
Iog10£    =0-43429448 
log,  10  =2-30258509 

TABLE  II. 


TABLE  III. 


No. 

Square  root. 

Cube  root. 

2 

1-4142136 

1-2599210 

3 

17320508 

1-4422496 

4 

2-0000000 

1-5874011 

5 

2-2360680 

1-7099759 

6 

2-4494897 

1-8171206 

7 

2-6457513 

1-9129312 

8 

2-8284271 

2-0000000 

9 

3-0000000 

2-0800837 

10 

3-1622777 

2-1544347 

11 

3-3166248 

2-2239801 

12 

3-4641016 

2-2894286 

13 

3-6055513 

2-3513347 

14 

3-7416574 

2-4101422 

15 

3-8729833 

2-4662121 

16 

4-0000000 

2-5198421 

17 

4-1231056 

2-5712816 

18 

4-2426407 

2-6207414 

19 

4-3588989 

2-6684016 

20 

4-4721360 

2-7144177 

21 

4-5825757 

2-7589243 

22 

4-6904158 

2-8020393 

23 

4-7958315 

2-8438670 

24 

4-8989795 

2-8844991 

25 

5-0000000 

2-9240177 

26 

5-0990195 

2-9624960 

27 

5-1961524 

3-0000000 

28 

5-2915026 

3-0365889 

29 

5-3851648 

3-0723168 

30 

5-4772256 

3-1072325 

N. 

logio  N. 

log*  N. 

2 

•3010300 

•69314718 

3 

•4771213 

1-09861229 

5 

•6989700 

1-60943791 

7 

•8450980 

1-94591015 

11 

1-0413927 

2-39789527 

13 

1-1139434 

2-56494936 

17 

1-2304489 

2-83321334 

19 

1-2787536 

2-94443898 

23 

1-3617278 

3-13549422 

29 

1-4623980 

3-36729583 

31 

1-4913617 

3-43398720 

37 

1-5682017 

3-61091791 

41 

1-6127839 

3-71357207 

43 

1-6334685 

3-76120012 

47 

1-6720979 

3-85014760 

53 

1-7242759 

3-97029191 

59 

1-7708520 

4-07753744 

61 

1-7853298 

4-11087386 

67 

1-8260748 

4-20469262 

71 

1-8512583 

4-26267988 

73 

1-8633229 

4-29045944 

79 

1-8976271 

4-36944785 

83 

1-9190781 

4-41884061 

89 

1-9493900 

4-48863637 

97 

1-9867717 

4-57471098 

101 

2-0043214 

4-61512052 

103 

2-0128372 

4-63472899 

107 

2-0293838 

4-67282883 

109 

2-0374265 

4-69134788 

•.*»• 

TABLES  OP  LOGARITHMS. 


479 


TABLE  IV. —  The  Natural  Logarithms  (each  increased  by  10) 
of  Numbers  between  0.00  and  0.99. 


N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

0.0 

5.395 

6.088 

6.493 

6.781 

7.004 

7.187 

7.341 

7.474 

7.592 

0.1 

7.697 

7.793 

7.880 

7.960 

8.034 

8.103 

8.167 

8.228 

8.285 

8.339 

0.2 

8.391 

8.439 

8.846 

8.530 

8.573 

8.614 

8.653 

8.691 

8.727 

8.762 

0.3 

8.796 

8.829 

8.861 

8.891 

8.921 

8.950 

8.978 

9.006 

9.032 

9.058 

0.4 

9.084 

9.108 

9.132 

9.156 

9.179 

9.201 

9.223 

9.245 

9.266 

9.287 

0.5 

9.307 

9.327 

9.346 

9.365 

9.384 

9.402 

9.420 

9.438 

9.455 

9.472 

0.6 

9.489 

9.506 

9.522 

9.538 

9.554 

9.569 

9.584 

9.600 

9.614 

9.629 

0.7 

9.643 

9.658 

9.671 

9.685 

9.699 

9.712 

9.726 

9.739 

9.752 

9.764 

0.8 

9.777 

9.789 

9.802 

9.814 

9.826 

9.837 

9.849 

9.861 

9.872 

9.883 

0.9 

9.895 

9.906 

9.917 

9.927 

9.938 

9.949 

9.959 

9.970 

9.980 

9.990 

TABLE  V. —  The  Natural  Logarithms  of  Numbers  between 
1.0  and  9.9. 


N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

1 

0.000 

0.095 

0.182 

0.262 

0.336 

0.405 

0.470 

0.531 

0.588 

0.642 

2 

0.693 

0.742 

0.788 

0.833 

0.875 

0.916 

0.956 

0.993 

1.030 

1.065 

3 

1.099 

1.131 

1.163 

1.194 

1.224. 

1.253 

1.281 

1.308 

1.335 

1.361 

4 

1.386 

1.411 

1.435 

1.459 

1.482 

1.504 

1.526 

2.548 

1,569 

1.589 

5 

1.609 

1.629 

1.649 

1.668 

1.686 

1.705 

1.723 

1.740 

1.758 

1.775 

6 

1.792 

1.808 

1.825 

1.841 

1.856 

1.872 

1.887 

1.902 

1.917 

1.932 

7 

1.946 

1.960 

1.974 

1.988 

2.001 

2.015 

2.028 

2.041 

2.054 

2.067 

8 

2.079 

2.092 

2.104 

2.116 

2.128 

2.140 

2.152 

2.163 

2.175 

2.186 

9 

2.197 

2.208 

2.219 

2.230 

2.241 

2.251 

2.262 

2.272 

2.282 

2.293 

480 


FINKEL'S    SOLUTION    BOOK. 


TABLE  VI. —  The  Values  in  Circular  Measure  of  Angles  which 
are  given  in  Degrees  and  Minutes. 


V 

0.0003 

20' 

0.0058 

7° 

0.1222 

80° 

1.3963 

2' 

0.0006 

30' 

0.0087 

8° 

0.1396 

90° 

1.5708 

3' 

0.0009 

40' 

0.0116 

9° 

0.1571 

100° 

1.7453 

4' 

0.0012 

50' 

0.0145 

10° 

0.1745 

110° 

1.9199 

5' 

0.0015 

60'orl° 

0.0175 

20° 

0.3491 

120° 

2.0944 

6' 

0.0017 

2° 

0.0349 

30° 

0.5236 

130° 

2.2689 

7' 

0.0020 

3° 

0.0524 

40° 

0.6981 

140° 

2.4435 

& 

0.0023 

4° 

0.0698 

50° 

0.8727 

150° 

2.6180 

9' 

0.0026 

5° 

0.0873 

60° 

1.0472 

160° 

2.7925 

10' 

0.0029 

6° 

0.1047 

70° 

1.2217 

170° 

2.9671 

TABLE  VII. — Equivalents  of  Radians  in  Degrees,  Minutes,  and 
Seconds  of  Arc. 


Radians. 

Equivalents. 

Radians. 

Equivalents. 

0.0001 

0°       0'     20".6 

0.0600 

3°     26'     15".9 

0.0002 

0°      0'    41".3 

0.0700 

4°      0'    38".5 

0.0003 

0°       T    01".9 

0.0800 

4°     35'    01".2 

0.0004 

0°      1'    22".5 

0.0900 

5°      9'    23".8 

0.0005 

0°      1'    43".l 

0.1000 

5°    43'    46".5 

0.0006 

0°      2'    03".8 

0.2000 

11°    27'    33".0 

0.0007 

0°      2'    24".4 

0.3000 

17°    11'    19".4 

0.0008 

0°      2'    45".0 

0.4000 

22°     55'    05".9 

0.0009 

0°      3'    05".6 

0.5000 

28°     38'    52".4 

0.0010 

0°      3'    26".3 

0.6000 

34°    22'    38".9 

0.0020 

0°      6'    52".5 

0.7000 

40°      6'    25".4 

0.0030 

0°     10'     18".8 

0.8000 

45°     50'    11".8 

0.0040 

0°     13'    45'M 

0.9000 

51°     33'    58".3 

0.0050 

0°     17'    11".3 

1.0000 

57°     17'    44".8 

0.0060 

0°    20'    37".6 

2.0000 

114°     35'    29".6 

0.0070 

0°    24'    03".9 

3.0000 

171°     53V    14".4 

0.0080 

0°    27'    30".l 

4.0000 

229°     10'    59".2 

0.0090 

0°    30'    56".4 

5.0000 

286°     28'    44".0 

0.0100 

0°    34'    22".6 

6.0000 

343°    46'    28".8 

0.0200 

1°      8'    45".3 

7.0000 

401°      4'    13".6 

0.0300 

1°    43'    07".9 

8.0000 

458°     21'    58".4 

0.0400 

2°     17'    30".6 

9.0000 

515°     39'    43".3 

0.0500 

2°     51'    53V2 

10.0000 

572°    57'    28".l 

EXAMPLE. 


481 


To  USE  TABLE  VII.— For  example,  express  1.3245  radians  in 
degrees,  minutes,  and  seconds. 


1.  1.0000  radian  ==  57° 

2.  .3000      "      =17° 

3.  .0200      "      =    1° 

4.  .0040      "      =    0° 

5.  .0005      "  0° 


17' 
11' 

8' 
13' 

V 


44".3 
19".4 
45".3 
45".l 
43".l 


Adding,  1.3245  radians  =  77°     53'    17".2 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 


LOAN  DEPT 


This  book  is  due  on  the  last  date  stamped  below,  or 

on  the  date  to  which  renewed. 
Renewed  books  are  subject  to  immediate  recall. 


LD  21A-50m-ll,'62 
(D3279slO)476B 


General  Library 

University  of  California 

Berkeley 


YC  22416 


